#### Please solve RD Sharma class 12 chapter Inverse Trigonometric Function exercise 3.9 question 1 subquestion (i) maths textbook solution

$\frac{24}{25}$

Hint:

Take the negative sign out of the sine inverse bracket and after that cosine of a negative angle will be equal to cosine of the same positive angle.

This way you get rid of the negative sign

Concept:

Inverse Trigonometry

Solution:

$cos\, cos((\frac{-7}{25}))=cos\, cos(-(\frac{7}{25}))$                                            $(-x)=-(x)$

$=cos\, cos((\frac{7}{25}))$                                            $[cos\, cos(-\theta )=cos\, \theta ]$

$Let\: \: \: sin^{-1}(\frac{7}{25})=x$                .......(1)

$sin\, x=\frac{7}{25}$

$cos\, cos\, x=\sqrt{1-x}$                                                                $[x+x=1]$

$=\sqrt{1-(\frac{7}{25})^{2}}$

$=\sqrt{1-\frac{49}{625}}$

$=\sqrt{\frac{625-49}{625}}$

$=\sqrt{\frac{576}{625}}=\frac{24}{25}$

$\therefore x=cos^{-1}(\frac{24}{25})$                                .......(2)

From (1) and (2)

$sin^{-1}(\frac{7}{25})=cos^{-1}(\frac{24}{25})$

$\therefore cos(sin^{-1}(\frac{7}{25}))=cos(cos^{-1}(\frac{24}{25}))$

$\therefore cos(sin^{-1}(\frac{7}{25}))=\frac{24}{25}$

Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer trigo ratio using identities.

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