#### Need solution for RD Sharma maths class 12 Chapter Inverse Trigonometric functions exercise 3.8 question 2 sub question (i)

Hint:

We convert in the form of $\tan (\alpha + \beta)$ so we know the formula of it.

Given:

We have to prove that $\tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\frac{17}{6}$

Solution:

LHS= $\tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)$

Let suppose that $\cos ^{-1}\frac{4}{5}=\alpha$ and $\tan ^{-1}\frac{2}{3}=\beta$,

\begin{aligned} &\cos \alpha=\frac{4}{5} \text { and } \tan \beta=\frac{2}{3} \\ &\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \end{aligned}

But we have not $\tan \alpha$. We have only $\tan \beta$. So, we convert $\cos \alpha$ into $\tan \alpha$.

$\cos \alpha =\frac{4}{5}=\frac{B}{H}$

Let, in $\Delta ABC$,  angle $CAB = \alpha$ and right angle at B.

\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(5)^{2}=4^{2}+(B C)^{2} \\ &(B C)^{2}=25-16 \\ \end{aligned}

$BC =\pm 9$

$BC = 3$

$\tan \alpha = \frac{3}{4}$

$[\because \tan \alpha = \frac{3}{4} \; and\: \tan \beta = \frac{2}{3} ]$

$\begin{gathered} \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ =\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}} \\ =\frac{\frac{9+8}{12-6}}{12}=\frac{17}{6}=\mathrm{RHS} \\ \tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\frac{17}{6} \end{gathered}$                                           [ we take LCM of 3,4]

Hence proved.