Find the values of each of the following:

    13. \tan \frac{1}{2}\left[\sin^{-1}\frac{2x}{1+x^2} + cos^{-1}\frac{1-y^2}{1+y^2} \right ],\;\;|x|<1,\;y>0 and xy<1

 

Answers (1)

Taking the value x = \tan \Theta  or  \tan^{-1}x = \Theta  and y = \tan \Theta  or  \tan^{-1} y = \Thetathen we have,

\tan \frac{1}{2}\left[\sin^{-1}\frac{2\tan \Theta}{1+(\tan \Theta)^2} + cos^{-1}\frac{1-\tan^2 \Theta}{1+(\tan \Theta)^2} \right ],

\tan \frac{1}{2}\left[\sin^{-1}(\sin2\Theta) + cos^{-1} (\cos 2\Theta) \right ]

\because \left[\cos^{-1}(\frac{1-\tan^2 \Theta}{1+ \tan^2\Theta}) = \cos^{-1} (\cos2 \Theta) , \right ]

\because \left[\sin^{-1}(\frac{2\tan\Theta}{1+ \tan^2\Theta}) = \sin^{-1} (\sin2 \Theta) \right ]

Then,

 =\tan \frac{1}{2}\left[2\tan^{-1}x + 2\tan^{-1}y \right ]  \because \left[\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right]

=\tan \left [ \tan^{-1}\frac{x+y}{1-xy} \right ]

 =\frac{x+y}{1-xy}  Ans.

 

 

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions