If { ^{ - 1}}{{2p}}{{1 + p2}} - { ^{ - 1}}{{1 - {q²}}{{1 + {q²}} = { ^{ - 1}}{{2x}}{{1 - {x²}}, then what is x equal to?

Correct Answer is : p−q1+pq{{p - q}}{{1 + pq}}1+pqp−q

Correct Answer is : p−q1+pq{{p - q}}{{1 + pq}}1+pqp−q

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UPSC NDA I 2019 Mathematics Solved Paper

Question : 40 of 120

Marks: +1, -0

{\sin ^{ - 1}}\frac{{2p}}{{1 + p2}} - {\cos ^{ - 1}}\frac{{1 - {q^2}}}{{1 + {q^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}

, then what is x equal to?

Solution:

Concept:

Use trigonometric identities to express inverse functions in simpler forms.

Explanation:

Given:

{\sin ^{ - 1}}\frac{{2p}}{{1 + {p^2}}} - {\cos ^{ - 1}}\frac{{1 - {q^2}}}{{1 + {q^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}

Let

p = \tan A

q = \tan B

, and

x = \tan C

Then

\sin^{-1}\frac{2\tan A}{1+\tan^2 A} = \sin^{-1}(\sin 2A) = 2A

(principal value).

Similarly,

\cos^{-1}\frac{1-\tan^2 B}{1+\tan^2 B} = \cos^{-1}(\cos 2B) = 2B

And

\tan^{-1}\frac{2\tan C}{1-\tan^2 C} = \tan^{-1}(\tan 2C) = 2C

Thus, the equation becomes

2A - 2B = 2C

, i.e.,

A - B = C

Taking

\tan

on both sides:

\tan(A - B) = \tan C = x

Using the identity

\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

Hence,

x = \frac{p - q}{1 + pq}

Answer:

\frac{{p - q}}{{1 + pq}}

(option B).

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