Question

prove that 1 minus sin theta + cos theta whole square is equal to 2 into 1 + cos theta into 1 minus sin theta

Answer 1

Answer:

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

Consider,

LHS

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

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

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

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

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

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

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

$=RHS$

Hence Proved

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

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

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

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

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

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

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

=RHS