Mathematics Questions and Answers – Trigonometric Identities – 2

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This set of Mathematics Aptitude Test for Class 10 focuses on “Trigonometric Identities – 2”.

1. If sec θ – tan θ = M then sec θ + tan θ = \(\frac {1}{M}\).
a) False
b) True
View Answer

Answer: b
Explanation: The appropriate trigonometric identity used here is sec2 θ – tan2 θ = 1.
(sec θ – tan θ) (sec θ + tan θ) = 1
M (sec θ + tan θ) = 1
(sec θ + tan θ) = \(\frac {1}{M}\)
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2. Find the correct trigonometric identity.
a) cos2 θ = 1 – sin2 θ
b) cos2 θ = 1 + sin2 θ
c) tan2 θ + sec2 θ = 1
d) tan2 θ = sec2 θ + 1
View Answer

Answer: a
Explanation: The appropriate trigonometric identity used here is sin2 θ + cos2 θ = 1.
cos2 θ = 1 – sin2 θ

3. Evaluate (sec θ – tan θ) (sec θ + tan θ).
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: b
Explanation: = (sec θ – tan θ) (sec θ + tan θ) = sec2 θ – tan2 θ
= 1
The identity used here is sec2 θ – tan2 θ = 1
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4. Evaluate \(\sqrt {\frac {1 – sin⁡ \, ⁡A}{1 + sin \, ⁡⁡A}}\).
a) cos A + tan A
b) cos A – tan A
c) tan A – cot A
d) sec A + tan A
View Answer

Answer: d
Explanation: \(\sqrt {\frac {1 – sin⁡ \, ⁡A}{1 + sin⁡ \, ⁡A}} = \sqrt {\frac {1 – sin⁡ \, ⁡A}{1 + sin \, ⁡⁡A}} . \sqrt {\frac {1 – sin \, ⁡A}{1 – sin \, ⁡⁡A}}\)
= \(\frac {\sqrt {(1 + sin \, ⁡A)^2}}{\sqrt {1 – sin^2} \, ⁡A} \)
= \(\frac {1 + sin⁡ \, ⁡A}{\sqrt {cos^2} \, ⁡A}\)     (∵ sin2 A + cos2 A = 1)
= \(\frac {1 + sin⁡ \, ⁡A}{cos⁡ \, ⁡A}\)
= sec A + tan A

5. Evaluate (cosec2 θ – cot2 θ)2 . (cosec θ + cot θ)2.
a) 1
b) 0
c) (cosec2 θ – cot2 θ)2
d) (cosec θ + cot θ)2
View Answer

Answer: d
Explanation: (cosec2 θ – cot2 θ)2 . (cosec θ + cot θ)2 = 1 (cosec θ + cot θ)2
= (cosec θ + cot θ)2
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6. (1 – sin2 A) (1 + tan2 A) equals to _____
a) – Sec2 θ Tan2 θ
b) – Sec2 θ Tan2 θ
c) 1
d) 0
View Answer

Answer: c
Explanation: (1 – sin2 A) (1 + tan2 A) = cos2 A . sec2 A
= cos2 A . \(\frac {1}{cos^2 A}\)
= 1

7. Evaluate (cosec A – 1) (cosec A + 1) (sec2 A – 1).
a) 0
b) 1
c) \(\frac {4}{3}\)
d) \(\frac {3}{4}\)
View Answer

Answer: b
Explanation: (cosec A – 1) (cosec A + 1) (sec2 A – 1) = (cosec2 A – 1) (sec2 A – 1)
= cot2A . tan2 A
= \(\frac {1}{tan^2 A}\) . tan2 A
= 1
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8. (sin A + cos A)2 is equal to _____
a) 1 + 2sin A cos A
b) 1 – 2sin A cos A
c) 2sin A cos A – 1
d) 2sin A cos A + 1
View Answer

Answer: a
Explanation: (sin A + cos A)2 = sin2 A + cos2 A + 2sin A cos A
= 1 + 2sin A cos A

9. Evaluate tan2 A + (1 + sec A) (sec A – 1).
a) 3 tan2 A
b) 0
c) 2tan2 A
d) 1
View Answer

Answer: c
Explanation: tan2 A + (1 + sec A) (sec A – 1) = tan2 A + (sec A + 1) (sec A – 1)
= tan2 A + (sec2 A – 1)
= tan2 A + tan2 A
= 2tan2 A
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10. (1 + cosec⁡ θ) (1 – cosec⁡ θ) + cot2⁡ θ is _____
a) Cot ⁡θ
b) 0
c) 1
d) Tan ⁡θ
View Answer

Answer: b
Explanation: (1 + cosec⁡ θ) (1 – cosec θ) + cot2 ⁡θ = (1 – cosec2 θ) + cot2 θ
= -cot2⁡ θ + cot2⁡ θ     (∵ cosec2 A – cot2 A = 1)
= 0

Sanfoundry Global Education & Learning Series – Mathematics – Class 10.

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