#### Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 3  Maths Textbook Solution.

Answer: $\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}} \text { then } x=\frac{a-b}{1+a b}$

Hints: First we will convert whole L.H.S part in $\tan ^{-1}$

Given:$\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$

Explanation:

As we know that

\begin{aligned} &2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}, 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ &2 \tan ^{1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}

Now $\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$

\begin{aligned} &2 \tan ^{-1} a-2 \tan ^{-1} b=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2\left(\tan ^{-1} a-\tan ^{-1} b\right)=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2 \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2 \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} x \\ &\frac{a-b}{1+a b}=\tan \left(\tan ^{-1} x\right) \\ &\frac{a-b}{1+a b}=x \\ &x=\frac{a-b}{1+a b} \end{aligned}

Hence it is proved that $x=\frac{a-b}{1+a b}$

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support