Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 11 Maths Textbook Solution.

Answer: $\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$

Given: For any a, b, x, y > 0

We have to prove that $\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$

where,$\alpha=-a x+b y \& \beta=b x+a y$

Hint:  First we will divide numerator and denominator of first function and second function by $b^{3}$ and $y^{3}$respectively, then use $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

Solution: Let us assume

\begin{aligned} L . H . S &=\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right) \\ &=\frac{2}{3} \tan ^{-1}\left[\frac{\frac{3 a b^{2}-a^{3}}{b^{3}}}{\frac{b^{3}-3 a^{2} b}{b^{3}}}\right]+\frac{2}{3} \tan ^{-1}\left[\frac{\frac{3 x y^{2}-x^{3}}{y^{3}}}{\frac{y^{3}-3 x^{2} y}{y^{3}}}\right] \end{aligned}

Dividing numerator and denominator of first function and second function by $b^{3}$ and $y^{3}$respectively.

$=\frac{2}{3} \tan ^{-1}\left[\frac{3\left(\frac{a}{b}\right)-\left(\frac{a}{b}\right)^{3}}{1-3\left(\frac{a}{b}\right)^{2}}\right]+\frac{2}{3} \tan ^{-1}\left[\frac{3\left(\frac{x}{y}\right)-\left(\frac{x}{y}\right)^{3}}{1-3\left(\frac{x}{y}\right)^{2}}\right]$

We know that,$3 \tan ^{-1} x=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$

$\Rightarrow \mathrm{LH} . \mathrm{S}=\frac{2}{3}\left[3 \tan ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\right]+\frac{2}{3}\left[3 \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\right]$

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

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

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

$=\tan ^{-1}\left(\frac{2 \times \frac{\beta}{\alpha}}{1-\left(\frac{\beta}{\alpha}\right)^{2}}\right)$                                                                               $\left[2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$

\begin{aligned} &=\tan ^{-1}\left(\frac{2 \beta}{\alpha} \times \frac{\alpha^{2}}{\alpha^{2}-\beta^{2}}\right) \\ &=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right) \\ &=\text { R.H.S } \end{aligned}

Hence,$\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$ Where $\alpha=-a x+b y \& \beta=b x+a y$

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