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

1.	(1-sin theta+cos theta) square prove that 2 (1+cos theta)(1-sin theta)
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 $= 1+sin^2\,\theta-2\times sin\,\theta+cos^2\,\theta+2cos\,\theta-2cos\,\theta\:sin\,\theta$ $= 1+sin^2\,\theta+cos^2\,\theta-2\times sin\,\theta+2cos\,\theta-2cos\,\theta\:sin\,\theta$ $= 1+1-2\times sin\,\theta+2cos\,\theta-2cos\,\theta\:sin\,\theta$ $= 2-2\times sin\,\theta+2cos\,\theta-2cos\,\theta\:sin\,\theta$ $= 2(1-sin\,\theta)+2cos\,\theta(1-sin\,\theta)$ $= (1-sin\,\theta)(2+2cos\,\theta)$ $= 2(1-sin\,\theta)$ = RHS Hence The relation is proved !!!!!!

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