Question

prove that (1-sintita+costita)square=2(1+costita)(1-sintita)​

Answer 1

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

$\boxed {\pink { (a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2ab+2bc+2ca }}$

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

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

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

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

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

$Hence \:proved .$

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