Question

solve this question plzz

oye sonu monu naam pata chal gaya bas karo yaar ab ​

Answer 1

Formula Used :

$\:$

${a}^{3} - {b}^{3} = (a - b)( {a}^{2} + {b}^{2} + ab)$

$\:$

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

$\:$

Solution :

$\:$

LHS

$\:$

$\sf \frac{ { \cos^{2}\theta }}{1 - \: tan\theta} + \frac{ {sin}^{3}\theta }{sin\theta \: - \: cos\theta}$

➙$\sf \frac{ {cos}^{2} \theta}{1 - \frac{sin\theta}{cos \theta} } + \frac{ {sin}^{3} \theta}{sin\theta - cos\theta}$

➙$\sf\frac{ {cos}^{2}\theta }{ \frac{cos\theta - sin\theta}{cos\theta} } + \frac{ {sin}^{3} \theta}{sin\theta - cos\theta}$

➙$\sf\frac{ {cos}^{3}\theta }{cos\theta - sin\theta} + \frac{ {sin}^{3}\theta }{sin\theta - cos\theta}$

➙$\sf\frac{ {cos}^{3}\theta }{cos\theta - sin\theta} + \frac{ {sin}^{3}\theta } { - (cos\theta - sin\theta)}$

➙$\sf\frac{ {cos}^{3}\theta }{cos\theta - sin\theta} - \frac{ {sin}^{3}\theta } { cos\theta - sin\theta}$

➙$\sf\frac{ {cos}^{3}\theta - {sin}^{3}\theta }{cos\theta - sin\theta}$

➙$\sf\frac{ {(cos}\theta - sin\theta)( {cos}^{2}\theta \: + \: {sin}^{2} \theta \: + \: sin \theta \: cos\theta) }{cos\theta - sin\theta}$

Cancelling $\sf cos\theta - sin\theta$

➙$\sf1 + sin\theta \: cos \theta$

RHS

------ ( Proved )