Question

1 + sin theta minus cos theta upon 1 + sin theta + cos theta ka whole square =1-cos theta upon 1+cos Theda. prove that

Answer 1

Answer:

Step-by-step explanation:

LHS

$(\frac{1+sin\theta -cos\theta}{1+sin\theta + cos\theta })^2$

Expending Numerator and Denominator using

(a + b + c )² = a² + b² + c² + 2ab + 2bc +2ca

$\frac{1+sin^2 \theta + cos^2 \theta + 2 sin\theta -2sin\theta cos\theta - 2 cos\theta}{1+sin^2\theta + cos^2\theta + 2 sin\theta +2sin\theta cos\theta +2 cos\theta}$

∵ sin² Ф + cos² Ф = 1

$\frac{1+ 1 + 2 sin\theta -2sin\theta cos\theta - 2 cos\theta}{1+1 + 2 sin\theta +2sin\theta cos\theta +2 cos\theta}$

$\frac{2 + 2 sin\theta -2sin\theta cos\theta - 2 cos\theta}{2 + 2 sin\theta +2sin\theta cos\theta +2 cos\theta}$

Taking 2 common from first two terms  and 2 cosФ from last two terms in both numerator and denominator

$\frac{2(1 + sin\theta) -2cos\theta(1+sin\theta)}{2(1 + sin\theta) +2cos\theta(1+sin\theta)}$

$\frac{(1+sin\theta)(2 - 2 cos\theta)}{(1+sin\theta)(2+2cos\theta)}$

$\frac{2(1-cos\theta)}{2(1+cos\theta)}$

$\frac{(1-cos\theta)}{(1+cos\theta)} = RHS$

Answer 2

Step-by-step explanation:

their is your answer. I hope it will help u.