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

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


We have

                \operatorname{cosec}\left(\cos ^{-1} \frac{3}{5}\right)


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

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

Let’s suppose that,

\begin{aligned} &\cos ^{-1} \frac{3}{5}=\alpha \\ &\cos \alpha=\frac{3}{5}=\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} \\ &5^{2}=3^{2}+(B C)^{2} \\ &25-9=(B C)^{2} \end{aligned}

BC =\pm 4                    [we will ignore the -ve sign because BC is a length and it can’t be -ve]

BC =4

\begin{aligned} &\cos e c \alpha=\frac{H}{P}=\frac{5}{4} \\ &\cos e c\left(\cos ^{-1} \frac{3}{5}\right)=\frac{5}{4} \end{aligned}



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