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Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 23 math

Answers (1)

Answer: 1

Given:  

\tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4}

Hint: Try to separate the \tan  function into RHS, So that the variables gets free.

Solution:

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

Now, 

\begin{array}{l} \tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4} \\\\ \tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\frac{\pi}{4} \\\\ \frac{x+y}{1-x y}=\tan \frac{\pi}{4} \\\\ \frac{x+y}{1-x y}=1 \\\\ x+y=1-x y \\\\ x+y+x y=1 \end{array}

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