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Provide solution for RD Sharma maths class 12  Chapter Inverse trigromtery functions exercise 3.8 question 1 sub question (I)

Answers (1)

\frac{7}{25}

Hint:

As we know that the value of \sin \left(\sin ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]

Given:

We have,

\sin \left(\sin ^{-1} \frac{7}{25}\right)

Solution:

So here in the expression already present the form of \sin \left(\sin ^{-1} x\right)

\sin \left(\sin ^{-1} \frac{7}{25}\right)=\frac{7}{25} \mid

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