Question

Prove the following identities:

`(1 + cot A + tan A) (sin A - cos A) = (sec A)/("cosec"^2 A) - ("cosec" A)/(sec^2 A) = sin A tan A - cot A cos A`

Answer 1

Given: `(1 + cot A + tan A) (sin A - cos A) = (sec A)/(cosec^2 A) - (cosec A)/(sec^2 A) = sin A tan A - cot A cos A`.

To Prove: `(1 + cot A + tan A) (sin A - cos A) = (sec A)/(cosec^2 A) - (cosec A)/(sec^2 A) = sin A tan A - cot A cos A`

Proof [Step-wise]:

1. Replace trig ratios by sin and cos where helpful:

`cot A = cos A/sin A`

`tan A = sin A/cos A`

`sec A = 1/cos A`

`"cosec" A = 1/sin A`

2. Simplify the left expression:

(1 + cot A + tan A)(sin A – cos A) 

= (1)(sin A – cos A) + cot A (sin A – cos A) + tan A (sin A – cos A) 

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

3. Evaluate each term:

`(cos A/sin A)(sin A - cos A) = cos A - (cos^2 A)/sin A`

`(sin A/cos A)(sin A - cos A) = (sin^2 A)/cos A - sin A`

4. Add the three results and cancel common terms:

`(sin A - cos A) + (cos A - (cos^2 A)/sin A) + ((sin^2 A)/cos A - sin A)` 

= `[sin A - sin A] + [-cos A + cos A] + ((sin^2 A)/cos A) - ((cos^2 A)/sin A)`

= `(sin^2 A)/cos A - (cos^2 A)/sin A`

5. Recognize this equals the third expression:

`(sin^2 A)/cos A - (cos^2 A)/sin A = sin A·tan A - cot A·cos A`

Because `sin A·tan A = sin A·(sin A/cos A) = (sin^2 A)/cos A` 

And `cot A·cos A = (cos A / sin A)·cos A = (cos^2 A)/sin A`

6. Now simplify the middle expression:

`(sec A)/(cosec^2 A) - (cosec A)/sec^2 A`

= `(1/cos A)/(1/sin^2 A) - (1/sin A)/(1/cos^2 A)` 

= `(1/cos A)·(sin^2 A) - (1/sin A)·(cos^2 A)`

= `(sin^2 A)/cos A - (cos^2 A)/sin A`

7. From steps 4–6 we have (1 + cot A + tan A)(sin A – cos A)

= `(sin^2 A)/cos A - (cos^2 A)/sin A` 

= `(sec A)/(cosec^2 A) - (cosec A)/(sec^2 A)`

= sin A·tan A – cot A·cos A

All three expressions are equal; hence the identity is proved.