Question

(1-sin theta+cos theta) square prove that 2 (1+cos theta)(1-sin theta)

Answer 1

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  

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

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

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

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

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

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

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

=  RHS

Hence The relation is proved !!!!!!