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If \( \Large sin \ \theta =\frac{a^{2}-1}{a^{2}+1} \) then the value of \( \Large sec \ \theta + tan \ \theta \) will be

A) \( \Large \frac{a}{\sqrt{2}} \)
B) \( \Large \frac{a}{a^{2}+1} \)
C) \( \Large \sqrt{2}a \)
D) a

Correct Answer:




D) a

Description for Correct answer:

\( \Large sin \theta =\frac{a^{2}-1}{a^{2}+1} \)

In, \( \Large \triangle ABC \)

\( \Large BC=\sqrt{AC^{2}-AB^{2}} \)

= \( \Large \sqrt{(a^{2}+1)^{2}-(a^{2}-1)^{2}} \)

= \( \Large \sqrt{a^{2}+1+2a^{2}-a^{2}-1+2a^{2}} \)

= \( \Large \sqrt{4a^{2}}=2a \)

\( \Large sec \theta +tan \theta =\frac{a^{2}+1}{2a}+\frac{a^{2}-1}{2a} \)

=\( \Large \frac{2a^{2}}{2a}=a \)

Part of solved Trigonometry questions and answers : >> Elementary Mathematics >> Trigonometry

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