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Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 32

Answers (1)

Answer:

$xy> -1$

Hint:

You must know the rules of inverse trigonometric function.

Given:

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

Solution:

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

Principal range of $\tan^{-1}$ is $\left ( \frac{-\pi}{2},\frac{\pi}{2} \right )$

Let $\tan^{-1}x=A$

$\tan^{-1}y=B$

So $\tan^{-1}\left ( \tan\left ( A-B \right ) \right ),A-B$ must be lie in $\left ( \frac{-\pi}{2},\frac{\pi}{2} \right )$

Now, if both $A,B< 0$ , then $A,B\: \epsilon \left ( \frac{-\pi}{2},0 \right )$

$A\: \epsilon \left ( \frac{-\pi}{2},0 \right )$ and $-B\: \epsilon \left ( 0,\frac{-\pi}{2} \right )$

$A-B\: \epsilon \left ( \frac{-\pi}{2},\frac{\pi}{2} \right )$

$\tan^{-1}\left ( \tan\left ( A-B \right ) \right )=A-B$

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

$\Rightarrow A-B< \frac{\pi}{2}$

$\Rightarrow A< \frac{\pi}{2}+B$

Apply $\tan$ ,

$\Rightarrow \tan A< \tan\left ( \frac{\pi}{2}+B \right )$

$\Rightarrow \tan\left ( \frac{\pi}{2}+\alpha \right )=-\cot\alpha$

$\Rightarrow \cot\alpha=\frac{-1}{\tan\alpha}$

So, $\tan A< \frac{-1}{\tan B}$

$\Rightarrow \tan A\tan B< -1$

$\Rightarrow \tan B< 0$

$\Rightarrow xy> -1$

Similarly,

$\Rightarrow \tan A \tan B> -1$

$\Rightarrow xy> -1$

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