#### Please solve RD Sharma Chapter Inverse trigonometric functions exercise 3.8 question 2 sub question (iii) maths text book solution

Hint:

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

Given:

We have to prove $\tan \left(\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}\right)=\frac{63}{16}$

Solution:

LHS = $\tan \left(\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}\right)$

$\sin ^{-1} \frac{5}{13}=\alpha \; and \; \cos ^{-1} \frac{3}{5}=\beta$

$\sin \alpha =\frac{5}{13} \; and \; \cos \beta = \frac{3}{5}$

$\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$                                                                                                                         …(i)

But we have not $\tan \alpha$ and  $\tan \beta$. So, we find out first then we put in equation (i)

$\sin \alpha = \frac{5}{13}=\frac{P}H{}$

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

\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(13)^{2}=AB^{2}+5^{2} \\ &(AB)^{2}=144 \\ \end{aligned}

$AB =\pm 12$                 [we will ignore the -ve sign because AB is a length and it can’t be -ve]

$AB =12$

$\therefore \tan \alpha = \frac{5}{12}$

Let, in $\Delta PQR$,  angle $RPQ = \beta$ and right angle at B.

$\cot \beta =\frac{3}{5}=\frac{B}{P}$

\begin{aligned} &\\ &(PR)^{2}=(QR)^{2}+(PQ)^{2} \\ &(5)^{2}=3^{2}+(QR)^{2} \\ &(QR)^{2}=16\\ \end{aligned}

$QR = \pm 4$

$QR = 4$

$\therefore \tan \beta = \frac{4}{3}$

Now,     \begin{aligned} \tan (\alpha+\beta)=& \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ &=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}} \end{aligned}                                                    [ put, $\tan \alpha = \frac{5}{12}$ and $\tan \beta = \frac{4}{3}$]

[ we take LCM of 3 and 12]

\begin{aligned} &=\frac{\frac{5+16}{12}}{\frac{36-20}{36}} \\ &=\frac{\frac{21}{12}}{\frac{16}{36}} \\ &=\frac{21}{12} \times \frac{36}{16} \\ &=\frac{21}{1} \times \frac{3}{16} \\ \tan (\alpha+\beta) &=\frac{63}{16}=\mathrm{RHS} \end{aligned}

Hence proved.