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Please solve RD Sharma Chapter Inverse trigonometric functions exercise 3.8 question 1 sub question (iv) maths text book solution

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As we know that the value of \left(\sin ^{-1} x\right) \text { is } x \text { for } x \in[-1,1] 


We have

                \sin \left(\sec ^{-1}\left(\frac{17}{8}\right)\right)


So here in place of \sin ^{-1} x  there  is \sec ^{-1} x.

So, we convert \sec ^{-1} x  into  \sin ^{-1} x

Let’s we suppose that,

\begin{aligned} &\sec ^{-1} \frac{17}{8}=\alpha \\ &\sec \alpha=\frac{17}{8} \end{aligned}

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

\begin{aligned} &\sec \alpha=\frac{H}{B}=\frac{17}{8} \\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(17)^{2}=8^{2}+(B C)^{2} \\ &(B C)^{2}=289-64 \end{aligned}

BC =\pm 15                            [we will ignore the -ve sign because BC is a length and it can’t be -ve]

BC = 15

\begin{aligned} &\sin \alpha=\frac{P}{H}=\frac{15}{17} \\ &\sin \left(\sec ^{-1}\left(\frac{17}{8}\right)\right)=\frac{15}{17} \end{aligned}



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