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Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 11

Answers (1)

Answer:

$\pi+\tan^{-1}\left ( \frac{x+y}{1-xy} \right )$

Hint:

You must know the values of inverse trigonometric function.

Given:

$x> 0,y> 0,xy> 1$ , then find $\tan^{-1}x+\tan^{-1}y$

Solution:

Using inverse trigonometry rule,

$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left ( \frac{x+y}{1-xy} \right ),xy> 1$

$\tan^{-1}x=a\Rightarrow x=\tan a$
$\tan^{-1}y=b\Rightarrow y=\tan b$

$\Rightarrow$ $0< a+b< \pi,\tan\left ( a+b \right )< 0$

$\frac{\pi}{2}< a+b< \pi$

$\frac{-\pi}{2}< a+b-\pi< 0$

$\Rightarrow$ $\tan\left ( a+b-\pi \right )=-\tan\left [ \pi-\left ( a+b \right ) \right ]$

$\Rightarrow$ $\tan\left ( a+b \right )=\left ( \frac{x+y}{1-xy} \right )$

$\Rightarrow$ $\tan^{-1}\left ( \frac{x+y}{1-xy} \right )=a+b -\pi$

$\Rightarrow$ $a+b= \pi+\tan^{-1}\left ( \frac{x+y}{1-xy} \right )$

$\Rightarrow$ $\tan^{-1}x+\tan^{-1}y= \pi+\tan^{-1}\left ( \frac{x+y}{1-xy} \right )$

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