#### Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 10 Maths Textbook Solution.

Answer: $\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$

Given: $\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)$where $\alpha=a x-b y \: \: \& \: \: \beta=a y+b x$

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

Solution: Let us assume

L.H.S:$=\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)$

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

\begin{aligned} &=\tan ^{-1}\left[\frac{\frac{2 a b}{a^{2}-b^{2}}+\frac{2 x y}{x^{2}-y^{2}}}{1-\left(\frac{2 a b}{a^{2}-b^{2}}\right)\left(\frac{2 x y}{x^{2}-y^{2}}\right)}\right] \\ &=\tan ^{-1}\left[\frac{2 a b x^{2}-2 a b y^{2}+2 x y a^{2}-2 x y b^{2}}{a^{2} x^{2}-a^{2} y^{2}-b^{2} x^{2}+b^{2} y^{2}-4 a b x y}\right] \\ &=\tan ^{-1}\left[\frac{2\left(a b x^{2}+x y a^{2}-a b y^{2}-x y b^{2}\right)}{a^{2} x^{2}-a^{2} y^{2}-b^{2} x^{2}+b^{2} y^{2}-4 a b x y}\right] \\ &=\tan ^{-1}\left[\frac{2[a x(b x+a y)-b y(a y+b x)]}{(a x-b y)^{2}-\left(a^{2} y^{2}+b^{2} x^{2}+2 a b x y\right)}\right] \end{aligned}

$=\tan ^{-1}\left[\frac{2(b x+a y)(a x-b y)}{(a x-b y)^{2}-(b x+a y)^{2}}\right]$                                                        $\left[(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$

$=\tan ^{-1}\left[\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right]$                                                                                        $[\alpha=a x-b y \& \beta=a y+b x]$

$=R.H.S$

Hence ,$\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$