Question

cos 2 theta/(1-tan theta)+sin 3 theta/(sin theta - cos theta)=1+sin theta.cos theta

Answer 1

LHS = cos²∅/(1 - tan∅) + sin³∅/(sin∅ - cos∅)

= cos²∅/(1 - sin∅/cos∅) + sin³∅/(sin∅ - cos∅)

= cos²∅.cos∅/(cos∅ - sin∅) + sin³∅/(sin∅ - cos∅)

= cos³∅/(cos∅ - sin∅) - sin³∅/(cos∅ - sin∅)

= (cos³∅ - sin³∅)/(cos∅ - sin∅)

= (cos∅ - sin∅)(sin²∅ + cos²∅ + sin∅.cos∅)/(cos∅ - sin∅)

= (sin²∅ + cos²∅ + sin∅.cos∅)

[ we know, Sin²x + cos²x = 1 so, sin²∅ + cos²∅ = 1 ]

= 1 + sin∅.cos∅ = RHS

Answer 2

Solution:

$\frac{ {cos}^{2} \theta}{1 - tan \: \theta} + \frac{ {sin}^{3} \theta}{sin \: \theta - cos \: \theta} \\ \\ as \: \: we \: \: know \: \: that \: \\ tan \: \theta = \frac{sin \: \theta}{cos \: \theta} \\ \\ = \frac{ {cos}^{2} \theta}{1 - \frac{sin \: \theta}{cos \: \theta} } + \frac{ {sin}^{3} \theta}{sin \: \theta - cos \: \theta} \\ \\ \\ = \frac{ {cos}^{2} \theta}{ \frac{cos \: \theta - sin \: \theta}{cos \: \theta} } + \frac{ {sin}^{3} \theta}{sin \: \theta - cos \: \theta} \\ \\ \\ = \frac{ {cos}^{3} \theta}{cos \: \theta - sin \: \theta} + \frac{ {sin}^{3} \theta}{sin \: \theta - cos \: \theta} \\ \\ \\ = \frac{ {cos}^{3} \theta}{cos \: \theta - sin \: \theta} - \frac{ {sin}^{3} \theta}{cos \: \theta - sin \: \theta}$

taking LCM

$\frac{ {cos}^{3} \theta - {sin}^{3} \theta}{cos \: \theta - sin \: \theta}$
As we know that

${x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ \\ so \: here \\ \\ = \frac{(cos \: \theta - sin \: \theta)( {cos}^{2}\theta + cos \: \theta \: sin \: \theta + {sin}^{2}\theta }{(cos \: \theta - sin \: \theta)}$
Because

${cos}^{2} x + {sin}^{2} x = 1 \\ so \\ = {cos}^{2} \theta+ {sin}^{2} \theta + cos \: \theta \: sin \: \theta \\ \\ = 1 + cos \: \theta \: sin \: \theta$

= R.H.S.

Hence proved