Question

1 - sin theta + cos theta whole bracket square is equal to 2 bracket 1 + cos theta bracket bracket 1 minus sin theta bracket

Answer 1

to prove: (1- sinθ+ cos θ)²=2(1+cosθ)(1-sinθ)
proof:
let (1- sin θ )= a
cosθ =b
LHS
(a+b)²=a²+b²+2ab
= 1+sin²θ-2sinθ+cos²θ+2ab
=1+1-2sinθ+2ab
=2(1-sinθ+ab)
=2(a+ab)
=2a(a+b)
RHS
2(1+b)(a)=2a+2ab=2a (a+b)
LHS =RHS

Answer 2

Answer:

To prove: $(1-sin\,\theta+cos\,\theta)^2=2(1+cos\,\theta)(1-sin\,\theta)$

Consider,

LHS = $(1-sin\,\theta+cos\,\theta)^2$

using ( a + b )² = a² + b² + 2ab, we get

       = $(1-sin\,\theta)^2+cos^2\,\theta+2\times cos\,\theta\times(1-sin\,\theta)$

Again using ( a - b )² = a² + b² - 2ab, we get

       = $1+sin^2\,\theta-2\times sin\,\theta+cos^2\,\theta+2cos\,\theta-2cos\,\theta\:sin\,\theta$

       = $1+sin^2\,\theta+cos^2\,\theta-2\times sin\,\theta+2cos\,\theta-2cos\,\theta\:sin\,\theta$

       = $1+1-2\times sin\,\theta+2cos\,\theta-2cos\,\theta\:sin\,\theta$

       = $2-2\times sin\,\theta+2cos\,\theta-2cos\,\theta\:sin\,\theta$

       = $2(1-sin\,\theta)+2cos\,\theta(1-sin\,\theta)$

       = $(1-sin\,\theta)(2+2cos\,\theta)$

       = $2(1-sin\,\theta)(1+cos\,\theta)$

       = RHS

Hence Proved