Prove the following trigonometric identities.

`(cot A - cos A)/(cot A + cos A) = (cosec A - 1)/(cosec A + 1)`

#### Solution

In the given question, we need to prove `(cot A - cos A)/(cot A + cos A) = (cosec A - 1)/(cosec A + 1)`

Here, we will first solve the LHS.

Now using `cot theta = (cos theta)/(sin theta)`, we get

`(cot A - cos A)/(cot A + cos A) = (cos A/sin A - cos A)/(cos A/sin A + cos A)`

`= ((cos A - cos Asin A)/sin A)/((cos A + cos A sin A)/sin A)`

On further solving by taking the reciprocal of the denominator, we get,

`((cos A - cos Asin A)/sin A)/((cos A + cos Asin A)/sin A) = ((cos A - cos AsinA)/sin A) (sin A/(cos A + cos A sin A))`

`= (cos A - cos AsinA)/(cos A + cos Asin A)`

Now, taking `cos A sin A` common from both the numerator and the denominator, we get

`(cos A - cos A sin A)/(cos A + cos Asin A) = (cos A sin A (1/sin A -1 ))/(cos A sin A (1/sin A + 1))`

`= ((1/sin A - 1))/((1/sin A + 1))`

`= (cosec A - 1)/(cosec A + 1)` `("using" 1/sin theta = cosec theta)`

Hence proved