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

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As we know that the range of \sec \left(\sec ^{-1} x\right) \text { is } x \text { for } x \in(-\infty,-1) \cup(1, \infty) 


We have

\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)


In place of  \sec ^{-1} x  there is \sin ^{-1} x

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

Let’s suppose that,

\begin{aligned} &\sin ^{-1} \frac{12}{13}=\alpha \\ &\sin \alpha=\frac{12}{13} \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} \\ &(13)^{2}=(AB)^{2}+(12)^{2} \\ &(AB)^2=25 \end{aligned}

AB = \pm 5                    [we will ignore the -ve sign because AB is a length and it can’t be -ve]

AB = 5

\begin{aligned} &\sec \alpha=\frac{13}{5} \\ &\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)=\sec \alpha=\frac{13}{5} \\ &\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)=\frac{13}{5} \end{aligned}



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