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Provide solution for RD Sharma maths class 12  Chapter Inverse trigromtery functions exercise 3.8 question 3 sub question (i)

Answers (1)

x= \pm \sqrt{\frac{35}{36}}


We know the value of\operatorname{cos}\left(\cos ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]


We have to solve \cos \left (\sin ^{-1} x \right )= \frac{1}{6}    to find the value of x.


Here, we have given \sin ^{-1} x in place of \cos ^{-1} x. So, we convert  \sin ^{-1} x into \cos ^{-1} x

Let’s suppose that,

            \sin ^{-1} x =\alpha

           \sin \alpha =x

We use the identity,                           

                \begin{aligned} &\sin ^{2} \alpha+\cos ^{2} \alpha=1 \\ &x^{2}+\cos ^{2} \alpha=1 \\ &\cos ^{2} \alpha=1-x^{2} \end{aligned}

                \cos \alpha =\pm \sqrt{1-x^2}                        [But we can eliminate negative value of \cos \alpha   , because in RHS of the question only positive value is given.]  

              \cos \alpha =\sqrt{1-x^2}                          …(i)                                                                    


 \begin{aligned} &\cos \left(\sin ^{-1} x\right)=\frac{1}{6} \\ &\cos \alpha=\frac{1}{6} \end{aligned}         .....(ii)


From equation (i) and (ii) we get,


Squaring on both sides,

                \begin{aligned} &\left(\sqrt{1-x^{2}}\right)^{2}=\left(\frac{1}{6}\right)^{2} \\ &1-x^{2}=\frac{1}{36} \\ &x^{2}=1-\frac{1}{36} \\ &x^{2}=\frac{35}{36} \\ &x=\pm \sqrt{\frac{35}{36}} \end{aligned}

 Therefore, the answer is x= \pm \sqrt{\frac{35}{36}} .

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