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Find the values of each of the expressions in Exercises 16 to 18.

    18. \tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )

Answers (1)


Given that \tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right ) 

we can take \sin^{-1}\frac{3}{5} = x,

then  \sin x = \frac{3}{5}

or  \cos x = \sqrt{1-\sin^{2}x}= \frac{4}{5}  

     \Rightarrow \tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}  

       \Rightarrow \tan^{-1}\frac{3}{4}= x

We have similarly;

\cot^{-1} \frac{3}{2} = \tan^{-1} \frac{2}{3}

Therefore we can write \tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )

=\tan\left(\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{2}{3} \right )

=\tan\left[\tan^{-1}\left ( \frac{\frac{3}{4}+\frac{2}{3}}{1- \frac{3}{4}.\frac{2}{3}} \right ) \right ]            from   As, \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]

=\tan \left (\tan^{-1} \frac{9+8}{12-6} \right ) = \tan \left (\tan^{-1} \frac{17}{6} \right )= \frac{17}{6}

Posted by

Divya Prakash Singh

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