Solve this

Question:

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

 

Solution:

$\mathrm{LHS}=\frac{\cos ^{3} \theta+\sin ^{3} \theta}{\cos \theta+\sin \theta}+\frac{\cos ^{3} \theta-\sin ^{3} \theta}{\cos \theta-\sin \theta}$

$=\frac{(\cos \theta+\sin \theta)\left(\cos ^{2} \theta-\cos \theta \sin \theta+\sin ^{2} \theta\right)}{(\cos \theta+\sin \theta)}+\frac{(\cos \theta-\sin \theta)\left(\cos ^{2} \theta+\cos \theta \sin \theta+\sin ^{2} \theta\right)}{(\cos \theta-\sin \theta)}$

$=\left(\cos ^{2} \theta+\sin ^{2} \theta-\cos \theta \sin \theta\right)+\left(\cos ^{2} \theta+\sin ^{2} \theta+\cos \theta \sin \theta\right)$

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

$=2$

$=$ RHS

Hence, LHS= RHS

 

Leave a comment