Question

COS(90+x).sec(-x).tan(180-x)/sec(360-x)sin(180+x)cot(90-x)

Answer 1

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Answer 2

Answer:

The value of

$\frac { \cos ( 90 + x ) \sec ( - x ) \tan ( 180 - x ) } { \sec ( 360 - x ) \sin ( 180 + x ) \cot ( 90 - x ) } = - 1$

Solution:

$\frac { \cos ( 90 + x ) \sec ( - x ) \tan ( 180 - x ) } { \sec ( 360 - x ) \sin ( 180 + x ) \cot ( 90 - x ) }$

We know that,

$\begin{aligned} \cos ( 90 + x ) & = - \sin x \\\\ \sec ( - x ) & = \sec x \\\\ \tan ( 180 - x ) & = - \tan x \\\\ \sec ( 360 - x ) & = \sec x \\\\ \sin ( 180 + x ) & = - \sin x \\\\ \cot ( 90 - x ) & = \tan x \end{aligned}$

$\frac { \cos ( 90 + x ) \sec ( - x ) \tan ( 180 - x ) } { \sec ( 360 - x ) \sin ( 180 + x ) \cot ( 90 - x ) } = \frac { - \sin x \times \sec x \times - \tan x } { \sec x \times - \sin x \times \tan x }$

=-1

Hence,

$\frac { \cos ( 90 + x ) \sec ( - x ) \tan ( 180 - x ) } { \sec ( 360 - x ) \sin ( 180 + x ) \cot ( 90 - x ) } = - 1$