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

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


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


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

Let’s suppose that  \sin ^{-1} \frac{3}{5}= \alphaand \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 \alphaand \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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