Question

1 - sin theta + cos theta ka whole square is equal to 2 bracket start 1 + cos theta bracket close then break it start 1 - sin theta bracket close to this identity

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Solution:

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

= $1^{2}+sin^{2}\theta+cos^{2}\theta\\+2\times1\times(-sin\theta)\\+2\times(-sin\theta)\times cos\theta\\+2\times cos\theta\times1$

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By algebraic identity,

(a+b+c)²

=a²+b²+c²+2ab+2bc+2ca

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

= $1+1+2cos\theta-2sin\theta-2sin\theta cos\theta$

\* By Trigonometric identity:*\

$\boxed {sin^{2}\theta+cos^{2}\theta=1}$

= $2+2cos\theta-2sin\theta-2sin\theta cos\theta$

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

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

= RHS

Therefore,.

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

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