Question

HELP PLEZ TRIGONOMETRY!

Answer 1

`(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha) = sin \alpha - cos \alpha`

Solution:

Given that we have to simplify:

`(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha)` ---- eqn 1

We know that,

`sin^2 x = 1 - cos^2 x`

Substitute the above identity in eqn 1

`(2\left(1-\cos ^(2) \alpha\right)-1)/(\sin \alpha+\cos \alpha)`

Simplify the above expression

`(2-2 \cos ^(2) \alpha-1)/(\sin \alpha+\cos \alpha)`

`(1-2 \cos ^(2) \alpha)/(\sin \alpha+\cos \alpha)` ------- eqn 2

By the trignometric identity,

`(sin x + cos x)(sin x - cos x) = 1-2cos^2 x`

Substitute the above identity in eqn 2

`((\sin \alpha+\cos \alpha)(\sin \alpha-\cos \alpha))/(\sin \alpha+\cos \alpha)`

Cancel the common factors in numerator and denominator

`((\sin \alpha+\cos \alpha)(\sin \alpha-\cos \alpha))/(\sin \alpha+\cos \alpha)=\sin \alpha-\cos \alpha`

Thus the simplified expression is:

`(2 \sin ^(2) \alpha-1)/(\sin \alpha+\cos \alpha) = sin \alpha - cos \alpha`