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Need solution for RD Sharma maths class 12 Chapter Inverse Trigonometric functions exercise 3.8 question 1 sub question (vii)

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As we know that the value of \tan \left(\tan ^{-1} x\right) \text { is } x \text { for } x \in \mathbb{R} \text { ( real number) }


We have

                \tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)


In place of \tan ^{-1} x  here is \cos ^{-1} x

So, we convert \cos ^{-1} x into \tan ^{-1} x.

Let’s suppose that,

\begin{aligned} &\cos ^{-1} \left (\frac{8}{17} \right )=\alpha \\ &\cos \alpha=\frac{8}{17}=\frac{B}{H} \end{aligned}

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


\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(17)^{2}=8^{2}+(B C)^{2} \\ &(B C)^{2}=289-64 \\ & (BC)^2= 25\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} &\tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)=\tan \alpha \\ &\tan \alpha=\frac{15}{8} \\ &\tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)=\frac{15}{8} \end{aligned}

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