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### Answers (1)

Hint:

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

Given:

We have to prove $\cos \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)=\frac{6}{5 \sqrt{13}}$

Solution:

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

Let’s suppose that  $\sin ^{-1} \frac{3}{5}= \alpha$and $\cot ^{-1} \frac{3}{2}=\beta$.

$\sin \alpha = \frac{3}{5} \; and\; \cot \beta =\frac{3}{2}$

$\cos\left ( \alpha +\beta \right )= \cos \alpha \cos \beta -\sin \alpha \sin \beta$

Here, we only have $\sin \alpha$ and $\cos \beta$.So we have to find $\cos \alpha$and $\sin \beta$.

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

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}=(AB)^{2}+3^{2} \\ &(AB )^{2}=25-9 \\ \end{aligned}

$AB=\pm 4$

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

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

$\cot \beta =\frac{3}{2}=\frac{B}{P}$

\begin{aligned} &\\ &(PR)^{2}=(QR)^{2}+(PQ)^{2} \\ &(PR)^{2}=2^{2}+3^{2} \\ &(PR)^{2}=4 + 9 \\ \end{aligned}

$PR =\pm \sqrt{13}$                        [we will ignore the -ve sign because PR is a length and it can’t be -ve]

$PR = \sqrt{13}$

\begin{aligned} &P R=\sqrt{13} \\ &\sin \beta=\frac{2}{\sqrt{13}}, \cos \beta=\frac{3}{\sqrt{13}} \\ &\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta \\ &=\frac{4}{5} \times \frac{3}{\sqrt{13}}-\frac{2}{\sqrt{13}} \times \frac{3}{5} \\ &=\frac{12}{5 \sqrt{13}}-\frac{6}{5 \sqrt{13}} \\ &\cos (\alpha+\beta)=\frac{6}{5 \sqrt{13}}=\mathrm{RHS} \end{aligned}

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