Question

Prove that (cosecA-sinA) (secA-cosA) = 1/tanA+cotA

Answer 1

this is the answer.if u like it mark it as brainliest answer.

Answer 2

Answer:

${ ( \csc A - \sin A ) ( \sec A - \cos A ) }= \frac{1}{\tan A + \cot A}$

Hence proved.

Solution:

$\begin{array} { c } { ( \csc A - \sin A ) ( \sec A - \cos A ) } \\\\ { = \left( \frac { 1 } { \sin A } - \sin A \right) \left( \frac { 1 } { \cos A } - \cos A \right) } \\\\ { \quad = \frac { 1 - \sin ^ { 2 } A } { \sin A } \left( \frac { 1 - \cos ^ { 2 } A } { \cos A } \right) } \\\\ { = \frac { \cos ^ { 2 } A } { \sin A } \left( \frac { \sin ^ { 2 } A } { \cos A } \right) } \\\\ { = \cos A \times \sin A } \end{array}$

$\begin{aligned} & = \frac { \cos A \times \sin A } { 1 } \\\\ & = \frac { \cos A \times \sin A } { \cos ^ { 2 } A + \sin ^ { 2 } A } \end{aligned}$

$= \frac { 1 } { \frac { \cos ^ { 2 } A + \sin ^ { 2 } A } { \cos A \times \sin A } }$

$= \frac { 1 } { \left( \frac { \cos ^ { 2 } A } { \cos A \sin A } \right) + \left( \frac { \sin ^ { 2 } A } { \cos A \sin A } \right) }$

$= \frac { 1 } { \left( \frac { \cos A } { \sin A } \right) + \left( \frac { \sin A } { \cos A } \right) }$

$= \frac{1}{\tan A + \cot A}$

Hence

${ ( \csc A - \sin A ) ( \sec A - \cos A ) }= \frac{1}{\tan A + \cot A}$

Hence proved