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

Hint:

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

Given:

We have to prove $\sin \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)=\frac{63}{65}$

Solution:

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

Let’s suppose that $\cos ^{-1} \frac{3}{5}=\alpha \; and \; \sin ^{-1} \frac{5}{13}=\beta$

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

\begin{aligned} &\Rightarrow \quad \sin \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right) \\ &\Rightarrow \quad \sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta \end{aligned}

…(i)

Here we don’t have  $\sin \alpha . \cos \beta$. So, we find out it and then put all the values in equation (i)

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

$\cos \alpha = \frac{3}{5}=\frac{B}H{}$

\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(5)^{2}=(3)^{2}+(BC)^{2} \\ &25^{2}=9^2+(BC)^2 \\ & (BC)^2 =16 \end{aligned}

$BC=\pm 4$

$BC= 4$

$\therefore \sin \alpha =\frac{4}{5}$

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

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

\begin{aligned} &\\ &(PR)^{2}=(PQ)^{2}+(QR)^{2} \\ &(13)^{2}=(PQ)^{2} + 5^2\\ &169=(PQ)^2+25\\ & (PQ)^2 =144 \end{aligned}

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

$PQ= 12$

$\therefore \cos \beta =\frac{12}{13}$

Now,    \begin{aligned} \sin (\alpha+\beta) &=\sin \alpha \cos \beta+\cos \alpha \sin \beta \\ &=\frac{4}{5} \times \frac{12}{13}+\frac{3}{5} \times \frac{5}{13} \\ &=\frac{48}{65}+\frac{15}{65} \\ \sin (\alpha+\beta) &=\frac{63}{65}=\mathrm{RHS} \end{aligned}

Hence proved.