Question

Pleaee solve this problem!!​

Answer 1

RHS=tanA/2

Explanation:

LHS=1+sinA-cosA/1+sinA+cosA

=(1-cosA)+sinA/(1+cos A)+sinA

=2sin^2A/2+2sinA/2*cosA/2

_____________________

2cos^2A/2+22sinA/2*cosA/2

=2sinA/2(sinA/2+cosA/2)

___________________

2cosA/2(sinA/2+cosA/2)

sinA/2

=_____

cosA/2

= tanA/2 proved.

Answer 2

Answer: see proof below

Explanation:

Use the following Double Angle Identities:

sin 2A = 2cos A · sin A

cos 2A = 2 cos²A - 1

Use the following Quotient Identity: tan A = (sin A)/(cos A)

Use the following Pythagorean Identity:

cos²A + sin²A = 1 --> sin²A = 1 - cos²A

Proof LHS → RHS

Given:
`(1+sin\theta - cos \theta)/(1+sin \theta +cos \theta)`

Let Ф = 2A:
`(1+sin2A - cos 2A)/(1+sin2A +cos2A)`

Un-factor:
`(\bigg((1- cos^2\ 2A)/(1+cos\ 2A)\bigg)+sin\ 2A )/(1+sin\ 2A +cos\ 2A)`

Pythagorean Identity:
`(\bigg((sin^2\ 2A)/(1+cos\ 2A)\bigg)+sin\ 2A )/(1+cos\ 2A +sin\ 2A)`

Simplify:
`(sin\ 2A)/(1+cos\ 2A)`

Double Angle Identity:
`(2sin\ A\cdot cos\ A)/(1+(2cos^2 A-1))`

Simplify:
`(2sin\ A\cdot cos\ A)/(2cos^2\ A)`

`=(2sin\ A\cdot cos\ A)/(2cos^2\ A)`

`=(sin\ A)/(cos\ A)`

Quotient Identity: tan A

`\text{Substitute} A = (\theta)/(2)}:\qquad tan(\theta)/(2)`

`tan(\theta)/(2) = tan(\theta)/(2)\quad \checkmark`