Solve - sin(sec^-1 (17/15)) If I just provide the answer, i.e., 15/17 instead of showing the complete solution, in CBSE board exam? If yes, then there is no need to provide the solution, and if not, kindly do so.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{No}, \mathrm{you} \mathrm{have} \mathrm{to} \mathrm{provide} \mathrm{complete} \mathrm{solution} \mathrm{otherwise} \mathrm{you} \mathrm{might} \mathrm{get} 0 \mathrm{marks}. \\ \mathrm{Also} \frac{15}{17} \mathrm{will} \mathrm{not} \mathrm{be} \mathrm{the} \mathrm{answer} \mathrm{here}. \\ \sin \left({\sec }^{-1}\frac{17}{15}\right) \\ \mathrm{Let} \\ {\sec }^{-1}\left(\frac{17}{15}\right)=\mathrm{A} \\ \mathrm{secA}=\frac{17}{15}=\frac{\mathrm{hypotenuse}\left(\mathrm{h}\right)}{\mathrm{base}\left(\mathrm{b}\right)} \\ \mathrm{Pependicular} \mathrm{p} \mathrm{is} \mathrm{given} \mathrm{by}: \\ {\mathrm{h}}^2={\mathrm{p}}^2+{\mathrm{b}}^2 \\ \Rightarrow {17}^2={\mathrm{p}}^2+{15}^2 \\ \Rightarrow 289={\mathrm{p}}^2+225 \\ \Rightarrow {\mathrm{p}}^2=289-225 \\ \Rightarrow {\mathrm{p}}^2=64 \\ \Rightarrow \mathrm{p}=8 \\ \mathrm{sinA}=\frac{\mathrm{p}}{\mathrm{h}} \\ \Rightarrow \mathrm{sinA}=\frac{8}{17} \\ \Rightarrow \mathrm{A}={\sin }^{-1}\left(\frac{8}{17}\right)={\sec }^{-1}\left(\frac{17}{15}\right) \\ \sin \left({\sec }^{-1}\frac{17}{15}\right)=\sin \left({\sin }^{-1}\left(\frac{8}{17}\right)\right)=\frac{\mathbf{8}}{\mathbf{17}}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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