#### Provide solution for RD Sharma maths class 12 chapter Inverse Trigonometric Function exercise 3.9 question 2 subquestion (iii)

$\frac{4}{5}$

Hint:

Get rid of the negative sign by using the property

$tan^{-1}(-x)=-tan^{-1}(x)$

Now use this property to simplify

$cos(-x)=cos(x)$

Now find the sec-1 equivalent of the given tan-1 term

Now use

$cos(x)=\frac{1}{sec (x)}$$cos(x)=\frac{1}{sec (x)}.\: \: to \; solve\: the\: sum$

Concept:

Inverse Trigonometry

Solution:

Let

$tan^{-1}(\frac{-3}{4})=x$

$tan\, x=\frac{-3}{4}$

$sec^{2}x=1+tan^{2}x$

$sec\, x=\sqrt{1+tan^{2}x}$

$=\sqrt{1+(\frac{-3}{4})^{2}}$

$=\sqrt{1+\frac{9}{16}}\; \; \: =\sqrt{\frac{25}{16}}$

$\therefore sec(x)=\frac{5}{4}$

$\therefore cos(x)=\frac{1}{sec(x)}\; \; \; =\frac{4}{5}$

$But \; \; x=tan^{-1}(\frac{-3}{4})$

$cos(tan^{-1}(\frac{-3}{4}))=\frac{4}{5}$

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 ratio using identities