Introduction to Trigonometry: 1 mark Questions (CBSE Class 10)
139 different 1 mark questions on Introduction to Trigonometry from CBSE Class 10 Maths board exams 2022–2026, newest first.
The value of sin 9 0 ∘ cos 9 0 ∘ − sin 2 6 0 ∘ is :
(A) 4 1 (B) 4 3 (C) − 4 3 (D) 4 5
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Answer: (C) − 4 3
sin 9 0 ∘ = 1 , cos 9 0 ∘ = 0 , sin 6 0 ∘ = 2 3 .Value = 1 × 0 − 4 3 = − 4 3 .
If sin θ = 7 1 , then tan θ is :
(A) 4 3 1 (B) 2 3 1 (C) 7 4 3 (D) 7 6
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cos θ = 1 − 49 1 = 7 48 = 7 4 3 .tan θ = c o s θ s i n θ = 4 3 1 .
The value of tan 3 0 ∘ tan 6 0 ∘ − sin 9 0 ∘ cos 9 0 ∘ is :
(A) 1(B) 0(C) 2(D) not defined
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Answer: (A) 1
tan 3 0 ∘ tan 6 0 ∘ = 3 1 × 3 = 1 .sin 9 0 ∘ cos 9 0 ∘ = 1 × 0 = 0 .Value = 1 – 0 = 1.
If sec θ = 3 , then the value of tan θ is :
(A) 3 − 1 (B) 6 0 ∘ (C) 2(D) 2
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tan 2 θ = sec 2 θ − 1 = 3 − 1 = 2 .tan θ = 2 .
The value of 2 tan 4 5 ∘ sin 2 6 0 ∘ − cos 9 0 ∘ is :
(A) − 4 1 (B) 2 3 (C) 4 3 (D) 4 1
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Answer: (B) 2 3
tan 4 5 ∘ = 1 , sin 2 6 0 ∘ = 4 3 , cos 9 0 ∘ = 0 .Value = 2 × 1 × 4 3 − 0 = 2 3 .
If value of cot θ is 5 , then sin θ equals
(A) 6 1 (B) 6 (C) 6 5 (D) 2 1
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cosec 2 θ = 1 + cot 2 θ = 1 + 5 = 6 cosec θ = 6 , so sin θ = 6 1
Assertion (A) : For an acute angle θ , cos θ is always less than 1. Reason (R) : In a right-angled triangle, hypotenuse is the longest side and cos θ = Hypotenuse Base .
(A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
In a right-angled triangle the hypotenuse is the longest side, so Base < Hypotenuse. R is true. Hence for an acute angle θ , cos θ = Hypotenuse Base < 1 . A is true and follows from R.
If sin θ = 11 1 , then cot θ equals
(A) 10 11 (B) 11 10 (C) 10 (D) 11
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cos θ = 1 − 11 1 = 11 10 cot θ = s i n θ c o s θ = 1/ 11 10 / 11 = 10
Given that sin θ = b a , then cos θ is equal to :
(A) b 2 − a 2 b (B) a b (C) b b 2 − a 2 (D) b 2 − a 2 a
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Take a right triangle with opposite side a and hypotenuse b . Adjacent side = b 2 − a 2 . cos θ = b b 2 − a 2 .
If cos A = 2 1 , then the value of sin 2 A + 2 cos 2 A is :
(A) 2 3 (B) 4 5 (C) − 1 (D) 2 1
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Answer: (B) 4 5
sin 2 A + 2 cos 2 A = ( sin 2 A + cos 2 A ) + cos 2 A = 1 + cos 2 A .= 1 + 4 1 = 4 5 .
If cos y = 0 , then what is the value of 2 1 cos 2 y ?
(A) 0(B) 2 1 (C) 2 1 (D) 2 2 1
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cos y = 0 gives y = 9 0 ∘ .2 1 cos 4 5 ∘ = 2 1 × 2 1 = 2 2 1 .
If cos A = 5 4 , then the value of tan A is :
(A) 5 3 (B) 4 3 (C) 3 4 (D) 3 5
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Answer: (B) 4 3
sin A = 1 − 25 16 = 5 3 tan A = c o s A s i n A = 4 3
If 2 sin A = 1 , then the value of tan A + cot A is :
(A) 3 (B) 3 4 (C) 2 3 (D) 1
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sin A = 2 1 , so A = 3 0 ∘ .tan 3 0 ∘ + cot 3 0 ∘ = 3 1 + 3 = 3 4
Given cot θ = 3 , the value of cos θ is :
(A) 3 1 (B) 10 1 (C) 10 3 (D) 3 10
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cot θ = 1 3 : adjacent = 3, opposite = 1, hypotenuse = 10 .cos θ = 10 3
Given that sin 2 α = 2 3 , the value of sin 3 α is :
(A) 4 3 3 (B) 2 1 (C) 1(D) 4 3
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Answer: (C) 1
sin 2 α = 2 3 = sin 6 0 ∘ , so 2 α = 6 0 ∘ , α = 3 0 ∘ sin 3 α = sin 9 0 ∘ = 1
The value of ( 2 1 tan 2 4 5 ∘ − cos 2 6 0 ∘ ) is :
(A) 0(B) − 2 1 (C) 4 1 (D) − 4 1
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Answer: (C) 4 1
tan 4 5 ∘ = 1 , cos 6 0 ∘ = 2 1 2 1 × 1 − 4 1 = 4 1
Assertion (A) : tan 2 θ is not defined at θ = 4 5 ∘ . Reason (R) : sin 9 0 ∘ = cos 9 0 ∘ .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (B) Both A and R are true, but R is not the correct explanation of A.
At θ = 4 5 ∘ , tan 2 θ = tan 9 0 ∘ = c o s 9 0 ∘ s i n 9 0 ∘ = 0 1 , not defined: A is true sin 9 0 ∘ = 1 , cos 9 0 ∘ = 0 , so R is truetan 9 0 ∘ is undefined because cos 9 0 ∘ = 0 , not because sin 9 0 ∘ = cos 9 0 ∘ , so R does not explain A
The value of ( 3 1 cot 2 3 0 ∘ − 2 1 sec 2 6 0 ∘ ) is :
(A) – 1(B) – 2(C) 8 5 (D) 8 7
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Answer: (A) – 1
cot 3 0 ∘ = 3 , so cot 2 3 0 ∘ = 3 ; sec 6 0 ∘ = 2 , so sec 2 6 0 ∘ = 4 3 1 × 3 − 2 1 × 4 = 1 − 2 = − 1
When sin A = 3 1 , the value of cot A is
(A) 3 2 2 (B) 2 2 (C) 2 2 1 (D) 3
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cos A = 1 − 9 1 = 3 2 2 cot A = s i n A c o s A = 1/3 2 2 /3 = 2 2
1 + c o t 2 A 1 + t a n 2 A equals to :
(A) tan 2 A (B) − 1 (C) − tan 2 A (D) cot 2 A
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Answer: (A) tan 2 A
1 + c o t 2 A 1 + t a n 2 A = cosec 2 A s e c 2 A = c o s 2 A s i n 2 A = tan 2 A
If 2 tan A = 3 , then value of sec A equals
(A) 2 13 (B) 4 13 (C) 13 2 (D) 2 13
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tan A = 2 3 sec A = 1 + tan 2 A = 1 + 4 9 = 2 13
s i n 2 A s e c 2 A − 1 is same as
(A) cos 2 A (B) sec 2 A (C) − sec 2 A (D) cot 2 A
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Answer: (B) sec 2 A
s i n 2 A s e c 2 A − 1 = s i n 2 A t a n 2 A = c o s 2 A s i n 2 A s i n 2 A = sec 2 A
For an acute angle θ , if sin θ = 9 1 , then value of 9 cosec θ − 1 9 cosec θ + 1 is
(A) 0(B) 81 80 (C) 1(D) 80 82
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Answer: (D) 80 82
cosec θ = s i n θ 1 = 9 9 × 9 − 1 9 × 9 + 1 = 80 82
Simplest form of s e c 2 A − 1 s e c A is
(A) sin A (B) tan A (C) cosec A (D) cos A
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Answer: (C) cosec A
sec 2 A − 1 = tan A (for acute A)t a n A s e c A = s i n A / c o s A 1/ c o s A = s i n A 1 = cosec A
For an acute angle θ , if cos θ = 8 1 , then 8 s e c θ − 1 8 s e c θ + 1 equals
(A) 63 64 (B) 0(C) 63 65 (D) 1
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Answer: (C) 63 65
sec θ = c o s θ 1 = 8 8 × 8 − 1 8 × 8 + 1 = 63 65
Which of the following statements is false ?
(A) tan 4 5 ∘ = cot 4 5 ∘ (B) sin 9 0 ∘ = tan 4 5 ∘ (C) sin 3 0 ∘ = cos 3 0 ∘ (D) sin 4 5 ∘ = cos 4 5 ∘
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Answer: (C) sin 3 0 ∘ = cos 3 0 ∘
tan 4 5 ∘ = cot 4 5 ∘ = 1 ; sin 9 0 ∘ = tan 4 5 ∘ = 1 ; sin 4 5 ∘ = cos 4 5 ∘ = 2 1 .sin 3 0 ∘ = 2 1 but cos 3 0 ∘ = 2 3 , so (C) is false.
The value of ( tan 2 A − c o s 2 A 1 ) is :
(A) more than 1(B) 1(C) 0(D) − 1
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Answer: (D) − 1
c o s 2 A 1 = sec 2 A .tan 2 A − sec 2 A = − ( sec 2 A − tan 2 A ) = − 1 .
The value of ( cot 2 A − s i n 2 A 1 ) is :
(A) more than 1(B) 1(C) 0(D) − 1
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Answer: (D) − 1
s i n 2 A 1 = cosec 2 A .cot 2 A − cosec 2 A = − ( cosec 2 A − cot 2 A ) = − 1 .
The value of ( s e c 2 A 1 + cosec 2 A 1 ) is :
(A) more than 1(B) 1(C) 0(D) − 1
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Answer: (B) 1
s e c 2 A 1 + cosec 2 A 1 = cos 2 A + sin 2 A = 1 .
The value of 1 − t a n 2 6 0 ∘ 2 t a n 6 0 ∘ is :
(A) − 3 (B) 3 (C) − 3 1 (D) − 3
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tan 6 0 ∘ = 3 , so tan 2 6 0 ∘ = 3 .Value = 1 − 3 2 3 = − 2 2 3 = − 3 .
The value of ( sin 2 A + cos 2 A ) + ( sec 2 A − tan 2 A ) − ( cot 2 A − cosec 2 A ) is :
(A) 1 (B) − 1 (C) 3 (D) − 2
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Answer: (C) 3
sin 2 A + cos 2 A = 1 .sec 2 A − tan 2 A = 1 .cot 2 A − cosec 2 A = − 1 .Value = 1 + 1 − ( − 1 ) = 3 .
( sin θ + cos θ ) 2 + ( sin θ − cos θ ) 2 =
(A) 1 (B) 2 (C) 2 + 2 sin θ cos θ (D) 2 + 4 sin θ cos θ
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Answer: (B) 2
Expand: ( sin 2 θ + 2 sin θ cos θ + cos 2 θ ) + ( sin 2 θ − 2 sin θ cos θ + cos 2 θ ) . = 2 ( sin 2 θ + cos 2 θ ) = 2 .
( sec θ − cos θ ) 2 + sin 2 θ − tan 2 θ = ?
(A) 0 (B) 1 (C) 2 (D) 4
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Answer: (A) 0
( sec θ − cos θ ) 2 = sec 2 θ − 2 + cos 2 θ (since sec θ cos θ = 1 ).Expression = ( sec 2 θ − tan 2 θ ) + ( sin 2 θ + cos 2 θ ) − 2 = 1 + 1 − 2 = 0 .
The value of s i n 2 4 5 ∘ + c o s 2 4 5 ∘ s e c 2 3 0 ∘ + t a n 2 3 0 ∘ is :
(A) 1(B) 3 5 (C) 3 13 (D) 7
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Answer: (B) 3 5
sec 2 3 0 ∘ = 3 4 , tan 2 3 0 ∘ = 3 1 , so the numerator = 3 5 .sin 2 4 5 ∘ + cos 2 4 5 ∘ = 2 1 + 2 1 = 1 .Value = 3 5 .
Assertion (A) : For sin θ = 1 , cos θ must be 0. Reason (R) : sin 2 θ − cos 2 θ = 1 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (C) Assertion (A) is true, but Reason (R) is false.
A: sin θ = 1 gives θ = 9 0 ∘ , so cos θ = 0 . A is true. R: the identity is sin 2 θ + cos 2 θ = 1 ; sin 2 θ − cos 2 θ = 1 is not true in general (e.g. θ = 0 ∘ gives − 1 ). R is false.
The value of s i n 3 0 ∘ + c o s 6 0 ∘ c o t 2 A − cosec 2 A is :
(A) 1(B) − 1 (C) 1 + 3 2 (D) 1 + 3 − 2
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Answer: (B) − 1
cosec 2 A − cot 2 A = 1 , so the numerator cot 2 A − cosec 2 A = − 1 .Denominator = sin 3 0 ∘ + cos 6 0 ∘ = 2 1 + 2 1 = 1 . Value = 1 − 1 = − 1 .
Assertion (A) : For an acute angle θ , cot θ = 1 ⇒ cosec θ = 2 . Reason (R) : cosec 2 θ − cot 2 θ = 1 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (D) Assertion (A) is false, but Reason (R) is true.
A: cot θ = 1 gives cosec 2 θ = 1 + cot 2 θ = 2 , so cosec θ = 2 , not 2. A is false. R: cosec 2 θ − cot 2 θ = 1 is a standard identity. R is true.
The value of θ for which sin 2 θ = tan 4 5 ∘ is :
(A) 22. 5 ∘ (B) 3 0 ∘ (C) 4 5 ∘ (D) 9 0 ∘
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Answer: (C) 4 5 ∘
tan 4 5 ∘ = 1 , so sin 2 θ = 1 = sin 9 0 ∘ .2 θ = 9 0 ∘ ⇒ θ = 4 5 ∘ .
Assertion (A) : For an angle θ , sec θ = 1 ⇒ tan θ = 0 . Reason (R) : sec 2 θ + tan 2 θ = 1 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (C) Assertion (A) is true, but Reason (R) is false.
A: tan 2 θ = sec 2 θ − 1 = 1 − 1 = 0 , so tan θ = 0 . A is true. R: the identity is sec 2 θ − tan 2 θ = 1 ; sec 2 θ + tan 2 θ = 1 is false in general (e.g. θ = 4 5 ∘ gives 2 + 1 = 3 ). R is false.
If 2 sin θ = 1 , then cot θ × cosec θ is equal to :
(A) 2 1 (B) 2 2 1 (C) 2 (D) 2 1
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sin θ = 2 1 , so θ = 4 5 ∘ .cot 4 5 ∘ × cosec 4 5 ∘ = 1 × 2 = 2 .
If sec θ − tan θ = 2 , then sec θ + tan θ is equal to :
(A) 2 1 (B) 2 (C) 2 1 (D) 2
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Answer: (A) 2 1
( sec θ − tan θ ) ( sec θ + tan θ ) = sec 2 θ − tan 2 θ = 1 .So sec θ + tan θ = 2 1 .
If sin A = 3 2 , then cos A is equal to :
(A) 2 3 (B) 3 5 (C) 3 1 (D) 3 1
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cos A = 1 − sin 2 A = 1 − 9 4 = 9 5 = 3 5 .
If tan A = 2 1 , then sin A is equal to :
(A) 5 2 (B) 3 1 (C) 5 1 (D) 1
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Take opposite side 1 and adjacent side 2; hypotenuse = 1 + 4 = 5 . sin A = 5 1 .
If tan A = 1, then 3 sin A + cos A is equal to
(A) 4 2 (B) 4(C) 2 2 (D) 4 × 4 5 ∘
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tan A = 1 gives A = 4 5 ∘ .3 sin 4 5 ∘ + cos 4 5 ∘ = 2 3 + 2 1 = 2 4 = 2 2 .
Assertion (A) : In a right angle triangle ABC, ∠ B = 9 0 ∘ . Therefore the value of cos (A + C) is equal to 0. Reason (R) : A + B + C = 18 0 ∘ and cos 9 0 ∘ = 0 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
A + B + C = 18 0 ∘ and B = 9 0 ∘ , so A + C = 9 0 ∘ .cos ( A + C ) = cos 9 0 ∘ = 0 , so A is true.R is true and is exactly the reason used.
If sin A = cos A, then 1 + t a n A 1 − t a n A is equal to
(A) 1(B) –1(C) 0(D) not defined
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Answer: (C) 0
sin A = cos A gives tan A = c o s A s i n A = 1 .1 + 1 1 − 1 = 2 0 = 0 .
If 3 sin θ = cos θ , then value of θ is
(A) 3 (B) 6 0 ∘ (C) 3 1 (D) 3 0 ∘
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Answer: (D) 3 0 ∘
tan θ = c o s θ s i n θ = 3 1 .So θ = 3 0 ∘ .
In △ A B C , ∠ B = 9 0 ∘ . If A C A B = 2 1 , then cos C is equal to
(A) 2 3 (B) 2 1 (C) 2 3 (D) 3 1
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Since ∠ B = 9 0 ∘ , AC is the hypotenuse and AB is opposite to C. sin C = A C A B = 2 1 , so C = 3 0 ∘ .cos C = cos 3 0 ∘ = 2 3 .
If sin θ = 9 1 , then tan θ is equal to
(A) 4 5 1 (B) 9 4 5 (C) 8 1 (D) 4 5
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cos θ = 1 − 81 1 = 9 80 = 9 4 5 .tan θ = c o s θ s i n θ = 9 1 × 4 5 9 = 4 5 1 .
If θ is an acute angle and 7 + 4 sin θ = 9 , then the value of θ is :
(A) 9 0 ∘ (B) 3 0 ∘ (C) 4 5 ∘ (D) 6 0 ∘
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Answer: (B) 3 0 ∘
4 sin θ = 2 , so sin θ = 2 1 .Since θ is acute, θ = 3 0 ∘ .
The value of tan 2 θ − ( c o s θ 1 × sec θ ) is :
(A) 1 (B) 0 (C) − 1 (D) 2
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Answer: (C) − 1
c o s θ 1 × sec θ = sec θ × sec θ = sec 2 θ .tan 2 θ − sec 2 θ = − ( sec 2 θ − tan 2 θ ) = − 1 .
If α + β = 9 0 ∘ and α = 2 β , then cos 2 α + sin 2 β is equal to :
(A) 0 (B) 2 1 (C) 1 (D) 2
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Answer: (B) 2 1
2 β + β = 9 0 ∘ , so β = 3 0 ∘ and α = 6 0 ∘ .cos 2 6 0 ∘ + sin 2 3 0 ∘ = 4 1 + 4 1 = 2 1 .
The value of ( tan A cosec A ) 2 − ( sin A sec A ) 2 is :
(A) 0 (B) 1 (C) − 1 (D) 2
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Answer: (B) 1
tan A cosec A = c o s A s i n A × s i n A 1 = sec A .sin A sec A = c o s A s i n A = tan A .Value = sec 2 A − tan 2 A = 1 .
If 7 cos 2 θ + 3 sin 2 θ = 4 , then the value of θ is :
(A) 3 0 ∘ (B) 4 5 ∘ (C) 6 0 ∘ (D) 9 0 ∘
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Answer: (C) 6 0 ∘
7 cos 2 θ + 3 ( 1 − cos 2 θ ) = 4 4 cos 2 θ = 1 , so cos θ = 2 1 θ = 6 0 ∘
If sin 3 0 ∘ tan 4 5 ∘ = k s e c 6 0 ∘ , then the value of k is :
(A) 4(B) 3(C) 2(D) 1
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Answer: (A) 4
sin 3 0 ∘ tan 4 5 ∘ = 2 1 × 1 = 2 1 sec 6 0 ∘ = 2 , so k 2 = 2 1 k = 4
If sin ( α + β ) = 1 , then the value of sin ( 2 α + β ) is :
(A) 2 1 (B) 2 1 (C) 0(D) 1
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sin ( α + β ) = 1 = sin 9 0 ∘ , so α + β = 9 0 ∘ sin ( 2 α + β ) = sin 4 5 ∘ = 2 1
If sin θ = cos θ ( 0 ∘ < θ < 9 0 ∘ ) , then the value of sec θ ⋅ sin θ is :
(A) 2 1 (B) 2 (C) 0(D) 1
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Answer: (D) 1
sin θ = cos θ with 0 ∘ < θ < 9 0 ∘ gives θ = 4 5 ∘ sec θ ⋅ sin θ = c o s θ s i n θ = tan 4 5 ∘ = 1
If tan 3 θ = 3 , then 2 θ equals :
(A) 6 0 ∘ (B) 3 0 ∘ (C) 2 0 ∘ (D) 1 0 ∘
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Answer: (D) 1 0 ∘
tan 3 θ = 3 = tan 6 0 ∘ , so 3 θ = 6 0 ∘ .θ = 2 0 ∘ .2 θ = 1 0 ∘ .
( cot θ + tan θ ) equals :
(A) cosec θ sec θ (B) sin θ sec θ (C) cos θ tan θ (D) sin θ cos θ
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Answer: (A) cosec θ sec θ
cot θ + tan θ = s i n θ c o s θ + c o s θ s i n θ = s i n θ c o s θ c o s 2 θ + s i n 2 θ .= s i n θ c o s θ 1 = cosec θ sec θ .
If sin 4 θ = 2 3 , then 3 θ equals :
(A) 6 0 ∘ (B) 2 0 ∘ (C) 1 5 ∘ (D) 5 ∘
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Answer: (D) 5 ∘
sin 4 θ = 2 3 = sin 6 0 ∘ , so 4 θ = 6 0 ∘ .θ = 1 5 ∘ .3 θ = 5 ∘ .
1 − c o s 2 θ c o s θ is equal to :
(A) cot θ (B) cos θ (C) s i n θ c o s θ (D) tan θ
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Answer: (A) cot θ
1 − cos 2 θ = sin 2 θ , so 1 − cos 2 θ = sin θ (for acute θ ).s i n θ c o s θ = cot θ .
tan 2 A = 3 tan A is true, when the measure of ∠ A is :
(A) 9 0 ∘ (B) 6 0 ∘ (C) 4 5 ∘ (D) 3 0 ∘
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Answer: (D) 3 0 ∘
For A = 3 0 ∘ : tan 6 0 ∘ = 3 and 3 tan 3 0 ∘ = 3 × 3 1 = 3 . True. A = 4 5 ∘ : tan 9 0 ∘ is not defined; A = 6 0 ∘ : tan 12 0 ∘ = 3 3 ; A = 9 0 ∘ : tan 9 0 ∘ not defined.
Which of the following statements is true ?
(A) sin 2 0 ∘ > sin 7 0 ∘ (B) sin 2 0 ∘ > cos 2 0 ∘ (C) cos 2 0 ∘ > cos 7 0 ∘ (D) tan 2 0 ∘ > tan 7 0 ∘
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Answer: (C) cos 2 0 ∘ > cos 7 0 ∘
For acute angles, sin and tan increase and cos decreases as the angle increases. So sin 2 0 ∘ < sin 7 0 ∘ , tan 2 0 ∘ < tan 7 0 ∘ , and cos 2 0 ∘ > cos 7 0 ∘ . Also sin 2 0 ∘ = cos 7 0 ∘ < cos 2 0 ∘ .
Which of the following is a trigonometric identity ?
(A) sin 2 θ = 1 + cos 2 θ (B) cosec 2 θ + cot 2 θ = 1 (C) sec 2 θ = 1 + tan 2 θ (D) sin 2 θ = 2 sin θ
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Answer: (C) sec 2 θ = 1 + tan 2 θ
sin 2 θ = 1 − cos 2 θ , so (a) is false.cosec 2 θ − cot 2 θ = 1 , so (b) is false.sec 2 θ = 1 + tan 2 θ holds for all valid θ : (c) is an identity.sin 2 θ = 2 sin θ fails e.g. at θ = 6 0 ∘ , so (d) is false.
sec A = 2 cos A is true for A =
(A) 0 ∘ (B) 3 0 ∘ (C) 4 5 ∘ (D) 6 0 ∘
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Answer: (C) 4 5 ∘
c o s A 1 = 2 cos A ⇒ cos 2 A = 2 1 ⇒ cos A = 2 1 A = 4 5 ∘
1 + t a n 2 3 0 ∘ 1 − t a n 2 3 0 ∘ is equal to
(A) sin 6 0 ∘ (B) cos 6 0 ∘ (C) tan 6 0 ∘ (D) sec 6 0 ∘
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Answer: (B) cos 6 0 ∘
tan 2 3 0 ∘ = 3 1 1 + 3 1 1 − 3 1 = 4/3 2/3 = 2 1 = cos 6 0 ∘
Assertion (A) : For an acute angle θ , sin θ = 5 3 ⇒ cos θ = − 5 4 . Reason (R) : For any value of θ , ( 0 ∘ ≤ θ ≤ 9 0 ∘ ) sin 2 θ + cos 2 θ = 1 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (D) Assertion (A) is false, but Reason (R) is true.
cos 2 θ = 1 − 25 9 = 25 16 For an acute angle cos θ > 0 , so cos θ = 5 4 , not − 5 4 . A is false. sin 2 θ + cos 2 θ = 1 holds for 0 ∘ ≤ θ ≤ 9 0 ∘ . R is true.
The value of 1 − t a n 2 6 0 ∘ 2 t a n 6 0 ∘ is same as the value of
(A) − tan 3 0 ∘ (B) − tan 6 0 ∘ (C) 2 sin 6 0 ∘ (D) 2 cos 6 0 ∘
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Answer: (B) − tan 6 0 ∘
tan 6 0 ∘ = 3 1 − 3 2 3 = − 3 = − tan 6 0 ∘
Assertion (A) : For an acute angle θ , sec θ = 3 ⇒ tan θ = 2 2 . Reason (R) : sec 2 θ = 1 − tan 2 θ for all values of θ .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (C) Assertion (A) is true, but Reason (R) is false.
tan 2 θ = sec 2 θ − 1 = 9 − 1 = 8 ; for acute θ , tan θ = 2 2 . A is true.The correct identity is sec 2 θ = 1 + tan 2 θ , so R is false.
The value of ( 1 − 2 sin 2 6 0 ∘ ) is same as that of
(A) sin 3 0 ∘ (B) − sin 3 0 ∘ (C) cos 6 0 ∘ (D) − cos 3 0 ∘
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Answer: (B) − sin 3 0 ∘
sin 2 6 0 ∘ = 4 3 1 − 2 × 4 3 = − 2 1 = − sin 3 0 ∘
Assertion (A) : For an acute angle θ , value of cosec θ cannot be 2 1 . Reason (R) : cosec θ ≥ 1 for 0 ∘ ≤ θ ≤ 9 0 ∘
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
cosec θ = s i n θ 1 and 0 < sin θ ≤ 1 in this range, so cosec θ ≥ 1 . R is true.2 1 < 1 , so cosec θ cannot equal 2 1 . A is true, and R explains A.
If x ( 1 + t a n 2 3 0 ∘ 2 t a n 3 0 ∘ ) = y ( 1 − t a n 2 3 0 ∘ 2 t a n 3 0 ∘ ) , then x : y =
(A) 1 : 1(B) 1 : 2(C) 2 : 1(D) 4 : 1
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Answer: (C) 2 : 1
tan 3 0 ∘ = 3 1 , tan 2 3 0 ∘ = 3 1 4/3 2/ 3 = 2 3 and 2/3 2/ 3 = 3 x ⋅ 2 3 = y 3 ⇒ y x = 2 x : y = 2 : 1
In a right triangle ABC, right-angled at A, if sin B = 4 1 , then the value of sec B is
(A) 4(B) 4 15 (C) 15 (D) 15 4
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cos B = 1 − 16 1 = 4 15 (B is acute)sec B = c o s B 1 = 15 4
If x = cos 3 0 ∘ − sin 3 0 ∘ and y = tan 6 0 ∘ − cot 6 0 ∘ , then
(A) x = y (B) x > y (C) x < y (D) x > 1 , y < 1
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Answer: (C) x < y
x = 2 3 − 2 1 = 2 3 − 1 ≈ 0.37 y = 3 − 3 1 = 3 2 ≈ 1.15 So x < y .
If x = 2 sin 6 0 ∘ cos 6 0 ∘ and y = sin 2 3 0 ∘ − cos 2 3 0 ∘ and x 2 = k y 2 , the value of k is
(A) 3 (B) − 3 (C) 3(D) − 3
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Answer: (C) 3
x = 2 × 2 3 × 2 1 = 2 3 , so x 2 = 4 3 y = 4 1 − 4 3 = − 2 1 , so y 2 = 4 1 4 3 = k × 4 1 ⇒ k = 3
If sin θ = 3 1 , then sec θ is equal to :
(A) 3 2 2 (B) 2 2 3 (C) 3(D) 3 1
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cos θ = 1 − 9 1 = 3 2 2 .sec θ = c o s θ 1 = 2 2 3 .
For what value of θ , sin 2 θ + sin θ + cos 2 θ is equal to 2 ?
(A) 4 5 ∘ (B) 0 ∘ (C) 9 0 ∘ (D) 3 0 ∘
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Answer: (C) 9 0 ∘
sin 2 θ + cos 2 θ = 1 , so 1 + sin θ = 2 .sin θ = 1 , so θ = 9 0 ∘ .
If sin A = 5 3 , then value of cot A is :
(A) 4 3 (B) 3 4 (C) 5 4 (D) 4 5
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Answer: (B) 3 4
cos A = 1 − 25 9 = 5 4 cot A = s i n A c o s A = 3/5 4/5 = 3 4
1 + c o t 2 A 1 + t a n 2 A is equal to
(A) sec 2 A (B) − 1 (C) cot 2 A (D) tan 2 A
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Answer: (D) tan 2 A
1 + tan 2 A = sec 2 A and 1 + cot 2 A = cosec 2 A cosec 2 A s e c 2 A = c o s 2 A s i n 2 A = tan 2 A
1 − t a n 2 3 0 ∘ 2 t a n 3 0 ∘ is equal to
(A) cos 6 0 ∘ (B) sin 6 0 ∘ (C) tan 6 0 ∘ (D) sin 3 0 ∘
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Answer: (C) tan 6 0 ∘
tan 3 0 ∘ = 3 1 Numerator = 3 2 , denominator = 1 − 3 1 = 3 2 Value = 3 2 × 2 3 = 3 = tan 6 0 ∘
If sin 2 θ = 4 3 , then θ is
(A) 3 0 ∘ (B) 4 5 ∘ (C) 6 0 ∘ (D) 9 0 ∘
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Answer: (C) 6 0 ∘
sin θ = 2 3 (taking the acute angle)θ = 6 0 ∘
If 5 cos A − 4 = 0 , then the value of tan A is
(A) 4 3 (B) 3 4 (C) 5 3 (D) 5 4
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Answer: (A) 4 3
cos A = 5 4 , so sin A = 1 − 25 16 = 5 3 tan A = 4/5 3/5 = 4 3
1 − s i n 2 A cosec 2 A − c o t 2 A is equal to
(A) sin 2 A (B) cos 2 A (C) sec 2 A (D) tan 2 A
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Answer: (C) sec 2 A
cosec 2 A − cot 2 A = 1 and 1 − sin 2 A = cos 2 A Value = c o s 2 A 1 = sec 2 A
In a right-angled triangle ABC, ∠ A = 9 0 ∘ and A B = A C . The value of sin C is :
(A) 0(B) 2 3 (C) 2 1 (D) 2 1
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A B = A C , so ∠ B = ∠ C .∠ B + ∠ C = 9 0 ∘ , so ∠ C = 4 5 ∘ .sin C = sin 4 5 ∘ = 2 1 .
If cos A = 2 1 , then tan A is equal to
(A) 3 1 (B) 3 (C) 3(D) 2 3
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cos A = 2 1 gives A = 6 0 ∘ .tan A = tan 6 0 ∘ = 3 .
If sin A = 2 1 , then cot A is equal to
(A) 3 1 (B) 1(C) 3 (D) 3 2
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sin A = 2 1 gives A = 3 0 ∘ .cot A = cot 3 0 ∘ = 3 .
If cos θ = 2 1 then tan θ is equal to
(A) 2 1 (B) 0(C) 1(D) 2 + 1
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Answer: (C) 1
cos θ = 2 1 gives θ = 4 5 ∘ .tan θ = tan 4 5 ∘ = 1 .
( t a n 2 θ 1 − s i n 2 θ 1 ) is equal to :
(A) 1 (B) − 1 (C) sec 2 θ (D) sin 2 θ
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Answer: (B) − 1
t a n 2 θ 1 − s i n 2 θ 1 = cot 2 θ − cosec 2 θ .Since cosec 2 θ − cot 2 θ = 1 , the value is − 1 .
( c o t 2 θ 1 ) − ( c o s 2 θ 1 ) is equal to :
(A) 1 (B) − 1 (C) 0 (D) sec 2 θ
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Answer: (B) − 1
c o t 2 θ 1 − c o s 2 θ 1 = tan 2 θ − sec 2 θ .Since sec 2 θ − tan 2 θ = 1 , the value is − 1 .
If 5 tan A = 3 , then the value of cot A is :
(A) 5 3 (B) 3 5 (C) 4 3 (D) 5 4
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Answer: (B) 3 5
tan A = 5 3 , so cot A = t a n A 1 = 3 5 .
( c o t 2 θ 5 − c o s 2 θ 5 ) is equal to :
(A) 1(B) 5(C) − 5 (D) 0
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Answer: (C) − 5
c o t 2 θ 5 − c o s 2 θ 5 = 5 tan 2 θ − 5 sec 2 θ .= 5 ( tan 2 θ − sec 2 θ ) = 5 × ( − 1 ) = − 5 .
If tan 2 θ = 3 , where θ is an acute angle, then the value of θ is :
(A) 3 0 ∘ (B) 6 0 ∘ (C) 0 ∘ (D) 4 5 ∘
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Answer: (B) 6 0 ∘
tan θ = 3 (positive, since θ is acute).tan 6 0 ∘ = 3 , so θ = 6 0 ∘ .
If sin 2 θ = 2 1 , then the value of tan 2 θ is :
(A) 3 1 (B) 3 (C) 0(D) 1
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Answer: (D) 1
cos 2 θ = 1 − sin 2 θ = 1 − 2 1 = 2 1 .tan 2 θ = c o s 2 θ s i n 2 θ = 1/2 1/2 = 1 .
If x = 4 sin θ , y = 4 cos θ , then the value of ( x 2 + y 2 ) is :
(A) 4(B) 4 1 (C) 16 1 (D) 16
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Answer: (D) 16
x 2 + y 2 = 16 sin 2 θ + 16 cos 2 θ .= 16 ( sin 2 θ + cos 2 θ ) = 16 .
If sec θ − tan θ = m , then the value of sec θ + tan θ is :
(A) 1 − m 1 (B) m 2 − 1 (C) m 1 (D) − m
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Answer: (C) m 1
( sec θ − tan θ ) ( sec θ + tan θ ) = sec 2 θ − tan 2 θ = 1 .So sec θ + tan θ = m 1 .
If cos ( α + β ) = 0 , then value of cos ( 2 α + β ) is equal to :
(A) 2 1 (B) 2 1 (C) 0(D) 2
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cos ( α + β ) = 0 = cos 9 0 ∘ , so α + β = 9 0 ∘ .cos ( 2 α + β ) = cos 4 5 ∘ = 2 1 .
If sin A = 3 2 , then value of cot A is :
(A) 2 5 (B) 2 3 (C) 4 5 (D) 3 2
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cos A = 1 − 9 4 = 3 5 .cot A = s i n A c o s A = 2/3 5 /3 = 2 5 .
For θ = 3 0 ∘ , the value of ( 2 sin θ cos θ ) is :
(A) 1(B) 2 3 (C) 4 3 (D) 2 3
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2 sin 3 0 ∘ cos 3 0 ∘ = 2 × 2 1 × 2 3 = 2 3 .
If sin θ = cos θ , ( 0 ∘ < θ < 9 0 ∘ ) , then value of ( sec θ ⋅ sin θ ) is :
(A) 2 1 (B) 2 (C) 1(D) 0
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Answer: (C) 1
sin θ = cos θ gives tan θ = 1 , so θ = 4 5 ∘ .sec θ ⋅ sin θ = c o s θ s i n θ = tan θ = 1
Assertion (A) : If sin A = 3 1 ( 0 ∘ < A < 9 0 ∘ ) , then the value of cos A is 3 2 2 Reason (R) : For every angle θ , sin 2 θ + cos 2 θ = 1 .
(A) Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true. Reason (R) does not give correct explanation of (A).(C) Assertion (A) is true but Reason (R) is not true.(D) Assertion (A) is not true but Reason (R) is true.
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Answer: (A) Both Assertion (A) and Reason (R) are true. Reason (R) is the correct explanation of Assertion (A).
R is the basic identity, true for every angle. Using R: cos A = 1 − 9 1 = 9 8 = 3 2 2 (positive as A is acute). So A is true. A follows directly from R, so R is the correct explanation.
If cos θ = 2 3 and sin ϕ = 2 1 , then tan ( θ + ϕ ) is :
(A) 3 (B) 3 1 (C) 1(D) not defined
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For acute angles, cos θ = 2 3 ⇒ θ = 3 0 ∘ . sin ϕ = 2 1 ⇒ ϕ = 3 0 ∘ .tan ( θ + ϕ ) = tan 6 0 ∘ = 3 .
If sin α = 2 3 , cos β = 2 3 , then tan α ⋅ tan β is :
(A) 3 (B) 3 1 (C) 1(D) 0
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Answer: (C) 1
For acute angles, sin α = 2 3 ⇒ α = 6 0 ∘ and cos β = 2 3 ⇒ β = 3 0 ∘ . tan 6 0 ∘ ⋅ tan 3 0 ∘ = 3 × 3 1 = 1 .
If sin θ = 1 , then the value of 2 1 sin ( 2 θ ) is :
(A) 2 2 1 (B) 2 1 (C) 2 1 (D) 0
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sin θ = 1 gives θ = 9 0 ∘ , so 2 θ = 4 5 ∘ .2 1 sin 4 5 ∘ = 2 1 × 2 1 = 2 2 1 .
The value of θ for which 2 sin 2 θ = 2 1 ; 0 ∘ ≤ θ ≤ 9 0 ∘ is :
(A) 3 0 ∘ (B) 6 0 ∘ (C) 4 5 ∘ (D) 9 0 ∘
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Answer: (A) 3 0 ∘
sin 2 θ = 4 1 , so sin θ = 2 1 (positive in the given range).θ = 3 0 ∘ .
If 3 x = 2 sin A , 3 y = 2 cos A , then the value of x 2 + y 2 is :
(A) 36(B) 9(C) 6(D) 18
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Answer: (A) 36
x = 6 sin A and y = 6 cos A .x 2 + y 2 = 36 ( sin 2 A + cos 2 A ) = 36 .
If 4 sec θ − 5 = 0 , then the value of cot θ is :
(A) 4 3 (B) 5 4 (C) 3 5 (D) 3 4
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Answer: (D) 3 4
sec θ = 4 5 , so cos θ = 5 4 and sin θ = 1 − 25 16 = 5 3 .cot θ = s i n θ c o s θ = 3 4 .
If x = a cos θ and y = b sin θ , then the value of b 2 x 2 + a 2 y 2 is :
(A) a 2 b 2 (B) ab (C) a 4 b 4 (D) a 2 + b 2
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Answer: (A) a 2 b 2
b 2 x 2 + a 2 y 2 = a 2 b 2 cos 2 θ + a 2 b 2 sin 2 θ .= a 2 b 2 ( cos 2 θ + sin 2 θ ) = a 2 b 2 .
If 5 tan θ − 12 = 0 , then the value of sin θ is :
(A) 12 5 (B) 13 12 (C) 13 5 (D) 5 12
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Answer: (B) 13 12
tan θ = 5 12 : take opposite side 12, adjacent side 5.Hypotenuse = 144 + 25 = 13 , so sin θ = 13 12 .
If 2 cos θ = 1 , then the value of θ is
(A) 4 5 ∘ (B) 6 0 ∘ (C) 3 0 ∘ (D) 9 0 ∘
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Answer: (B) 6 0 ∘
2 cos θ = 1 gives cos θ = 2 1 .Since cos 6 0 ∘ = 2 1 , θ = 6 0 ∘ .
If 3 tan θ = 1 , then the value of θ is
(A) 3 0 ∘ (B) 4 5 ∘ (C) 6 0 ∘ (D) 9 0 ∘
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Answer: (A) 3 0 ∘
3 tan θ = 1 gives tan θ = 3 1 .Since tan 3 0 ∘ = 3 1 , θ = 3 0 ∘ .
9 sec 2 A − 9 tan 2 A is equal to :
(A) 9(B) 0(C) 8(D) 9 1
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Answer: (a) 9
9 sec 2 A − 9 tan 2 A = 9 ( sec 2 A − tan 2 A ) = 9 × 1 = 9
The value of 2 sin 2 3 0 ∘ + 3 tan 2 6 0 ∘ − cos 2 4 5 ∘ is :
(A) 3 3 (B) 2 19 (C) 4 9 (D) 9
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Answer: (d) 9
2 ( 2 1 ) 2 + 3 ( 3 ) 2 − ( 2 1 ) 2 = 2 1 + 9 − 2 1 = 9
If tan A = 5 2 , then the value of 1 − s i n 2 A 1 − c o s 2 A is :
(A) 4 25 (B) 25 4 (C) 5 4 (D) 4 5
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Answer: (b) 25 4
1 − s i n 2 A 1 − c o s 2 A = c o s 2 A s i n 2 A = tan 2 A = ( 5 2 ) 2 = 25 4
8 ( cos 2 A + sin 2 A ) is equal to :
(A) 1(B) 0(C) 9(D) 8
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Answer: (d) 8
cos 2 A + sin 2 A = 1 So 8 ( cos 2 A + sin 2 A ) = 8
The value of s e c 4 5 ∘ + t a n 4 5 ∘ s i n 9 0 ∘ + c o s 6 0 ∘ is :
(A) 1(B) 2 3 ( 2 + 1 ) (C) 2 3 ( 2 − 1 ) (D) 2 + 1 1 + 3
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2 + 1 1 + 2 1 = 2 ( 2 + 1 ) 3 Multiply numerator and denominator by 2 − 1 : 2 ( 2 − 1 ) 3 ( 2 − 1 ) = 2 3 ( 2 − 1 )
The value of 5 sin 2 9 0 ∘ − 2 cos 2 0 ∘ is :
(A) − 2 (B) 5(C) 3(D) − 3
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Answer: (C) 3
sin 9 0 ∘ = 1 , cos 0 ∘ = 1 .5 ( 1 ) − 2 ( 1 ) = 3 .
( 3 2 sin 0 ∘ − 5 4 cos 0 ∘ ) is equal to :
(A) 3 2 (B) 5 − 4 (C) 0(D) 15 − 2
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Answer: (B) 5 − 4
sin 0 ∘ = 0 , cos 0 ∘ = 1 .3 2 ( 0 ) − 5 4 ( 1 ) = − 5 4 .
( 3 sin 2 3 0 ∘ − 4 cos 2 6 0 ∘ ) is equal to :
(A) 4 5 (B) − 4 3 (C) − 4 1 (D) − 4 9
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Answer: (C) − 4 1
sin 3 0 ∘ = 2 1 , cos 6 0 ∘ = 2 1 .3 × 4 1 − 4 × 4 1 = 4 3 − 1 = − 4 1 .
If sin θ = b a , then sec θ is equal to ( 0 ≤ θ ≤ 9 0 ∘ ) :
(A) b 2 − a 2 a (B) b 2 − a 2 b (C) b b 2 − a 2 (D) a b 2 − a 2
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cos θ = 1 − b 2 a 2 = b b 2 − a 2 .sec θ = c o s θ 1 = b 2 − a 2 b .
If sin θ = 4 3 , then s i n θ ( s e c 2 θ − 1 ) c o s 2 θ equals :
(A) 5 3 (B) 4 3 (C) 3 4 (D) 16 9
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Answer: (B) 4 3
( sec 2 θ − 1 ) cos 2 θ = tan 2 θ cos 2 θ = sin 2 θ .So the expression = s i n θ s i n 2 θ = sin θ = 4 3 .
If cos θ = 7 3 , then 1 − s i n 2 θ c o s θ is equal to :
(A) 40 3 (B) 7 3 (C) 3 7 (D) 40 7
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Answer: (C) 3 7
1 − sin 2 θ = cos 2 θ .So the expression = c o s 2 θ c o s θ = c o s θ 1 = 3 7 .
In △ A B C , right angled at C, if tan A = 7 8 , then the value of cot B is
(A) 8 7 (B) 7 8 (C) 113 7 (D) 113 8
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Answer: (B) 7 8
With the right angle at C, tan A = A C B C and cot B = A C B C . So cot B = tan A = 7 8 .
Assertion (A) : For 0 < θ ≤ 9 0 ∘ , cosec θ − cot θ and cosec θ + cot θ are reciprocal of each other. Reason (R) : cot 2 θ − cosec 2 θ = 1
(A) Both Assertion (A) and Reason (R) are true; and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (C) Assertion (A) is true, but Reason (R) is false.
( cosec θ − cot θ ) ( cosec θ + cot θ ) = cosec 2 θ − cot 2 θ = 1 , so the two are reciprocals: A is true.But cot 2 θ − cosec 2 θ = − 1 , not 1: R is false.
In the given figure, △ P QR is a right triangle right angled at Q. If PQ = 4 cm and PR = 8 cm, then ∠ P is
(A) 6 0 ∘ (B) 4 5 ∘ (C) 3 0 ∘ (D) 1 5 ∘
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Answer: (A) 6 0 ∘
cos P = P R P Q = 8 4 = 2 1 .So ∠ P = 6 0 ∘ .
If 2 tan A = 3 , then the value of 4 s i n A − 3 c o s A 4 s i n A + 3 c o s A is
(A) 13 7 (B) 13 1 (C) 3(D) does not exist
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Answer: (C) 3
tan A = 2 3 .Divide numerator and denominator by cos A : 4 t a n A − 3 4 t a n A + 3 . = 6 − 3 6 + 3 = 3
[ 4 3 tan 2 3 0 ∘ − sec 2 4 5 ∘ + sin 2 6 0 ∘ ] is equal to
(A) –1(B) 6 5 (C) 2 − 3 (D) 6 1
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Answer: (A) –1
tan 2 3 0 ∘ = 3 1 , sec 2 4 5 ∘ = 2 , sin 2 6 0 ∘ = 4 3 .Value = 4 3 × 3 1 − 2 + 4 3 = 4 1 − 2 + 4 3 = − 1
[ 8 5 sec 2 6 0 ∘ − tan 2 6 0 ∘ + cos 2 4 5 ∘ ] is equal to
(A) 3 − 5 (B) 2 − 1 (C) 0(D) 4 − 1
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Answer: (C) 0
sec 2 6 0 ∘ = 4 , tan 2 6 0 ∘ = 3 , cos 2 4 5 ∘ = 2 1 .Value = 8 5 × 4 − 3 + 2 1 = 2 5 − 3 + 2 1 = 0
sec θ when expressed in terms of cot θ , is equal to :
(A) c o t θ 1 + c o t 2 θ (B) 1 + cot 2 θ (C) c o t θ 1 + c o t 2 θ (D) c o t θ 1 − c o t 2 θ
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Answer: (C)
c o t θ 1 + c o t 2 θ
sec 2 θ = 1 + tan 2 θ = 1 + c o t 2 θ 1 = c o t 2 θ c o t 2 θ + 1 sec θ = c o t θ 1 + c o t 2 θ
Which of the following is true for all values of θ ( 0 ∘ ≤ θ ≤ 9 0 ∘ ) ?
(A) cos 2 θ − sin 2 θ = 1 (B) cosec 2 θ − sec 2 θ = 1 (C) sec 2 θ − tan 2 θ = 1 (D) cot 2 θ − tan 2 θ = 1
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Answer: (C) sec 2 θ − tan 2 θ = 1
From the identity 1 + tan 2 θ = sec 2 θ we get sec 2 θ − tan 2 θ = 1 . The others fail, e.g. at θ = 4 5 ∘ options (A), (B), (D) all give 0.
( sec 2 θ − 1 ) ( cosec 2 θ − 1 ) is equal to :
(A) − 1 (B) 1(C) 0(D) 2
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Answer: (B) 1
sec 2 θ − 1 = tan 2 θ and cosec 2 θ − 1 = cot 2 θ Product = tan 2 θ cot 2 θ = 1
If tan θ = 12 5 , then the value of s i n θ − c o s θ s i n θ + c o s θ is :
(A) − 7 17 (B) 7 17 (C) 13 17 (D) − 13 7
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Answer: (A) − 7 17
Divide numerator and denominator by cos θ : t a n θ − 1 t a n θ + 1 = 12 5 − 1 12 5 + 1 = − 12 7 12 17 = − 7 17 .
( 1 + t a n 2 3 0 ∘ 2 t a n 3 0 ∘ ) is equal to :
(A) sin 6 0 ∘ (B) cos 6 0 ∘ (C) tan 6 0 ∘ (D) sin 3 0 ∘
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Answer: (A) sin 6 0 ∘
tan 3 0 ∘ = 3 1 , so 1 + tan 2 3 0 ∘ = 1 + 3 1 = 3 4 .4/3 2/ 3 = 3 2 × 4 3 = 2 3 = sin 6 0 ∘ .
( 1 + t a n 2 3 0 ∘ 1 − t a n 2 3 0 ∘ ) is equal to :
(A) sin 6 0 ∘ (B) cos 6 0 ∘ (C) tan 6 0 ∘ (D) cos 3 0 ∘
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Answer: (B) cos 6 0 ∘
tan 2 3 0 ∘ = 3 1 .1 + 3 1 1 − 3 1 = 4/3 2/3 = 2 1 = cos 6 0 ∘ .
If sec θ − tan θ = 3 1 , then the value of ( sec θ + tan θ ) is :
(A) 3 4 (B) 3 2 (C) 3 1 (D) 3
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Answer: (D) 3
( sec θ + tan θ ) ( sec θ − tan θ ) = sec 2 θ − tan 2 θ = 1 .So sec θ + tan θ = 1/3 1 = 3 .
s i n 2 θ c o s 2 θ − s i n 2 θ 1 , in simplified form, is :
(A) tan 2 θ (B) sec 2 θ (C) 1(D) –1
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Answer: (D) –1
s i n 2 θ c o s 2 θ − 1 = s i n 2 θ − s i n 2 θ = − 1
If θ is an acute angle of a right angled triangle, then which of the following equation is not true ?
(A) sin θ cot θ = cos θ (B) cos θ tan θ = sin θ (C) cosec 2 θ − cot 2 θ = 1 (D) tan 2 θ − sec 2 θ = 1
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Answer: (D) tan 2 θ − sec 2 θ = 1
sin θ ⋅ s i n θ c o s θ = cos θ is true.cos θ ⋅ c o s θ s i n θ = sin θ is true.cosec 2 θ − cot 2 θ = 1 is an identity.sec 2 θ − tan 2 θ = 1 , so tan 2 θ − sec 2 θ = − 1 , not 1. (D) is not true.
Statement A (Assertion) : For 0 < θ ≤ 9 0 ∘ , cosec θ − cot θ and cosec θ + cot θ are reciprocal of each other. Statement R (Reason) : cosec 2 θ − cot 2 θ = 1
(A) Both Assertion (A) and Reason (R) are true; and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true; but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true but Reason (R) is false.(D) Assertion (A) is false but Reason (R) is true.
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Answer: (A) Both A and R are true and R is the correct explanation of A.
R is a standard identity, so R is true. ( cosec θ − cot θ ) ( cosec θ + cot θ ) = cosec 2 θ − cot 2 θ = 1 .So the two expressions are reciprocals: A is true, and R explains A.
( cos 4 A − sin 4 A ) on simplification, gives
(A) 2 sin 2 A − 1 (B) 2 sin 2 A + 1 (C) 2 cos 2 A + 1 (D) 2 cos 2 A − 1
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Answer: (D) 2 cos 2 A − 1
cos 4 A − sin 4 A = ( cos 2 A − sin 2 A ) ( cos 2 A + sin 2 A ) = cos 2 A − sin 2 A = cos 2 A − ( 1 − cos 2 A ) = 2 cos 2 A − 1
If tan θ = y x , then cos θ is equal to
(A) x 2 + y 2 x (B) x 2 + y 2 y (C) x 2 − y 2 x (D) x 2 − y 2 y
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Take opposite side = x and adjacent side = y; hypotenuse = x 2 + y 2 . cos θ = hypotenuse adjacent = x 2 + y 2 y
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