Question

prove that sin^6 theta + cos^6 theta + 3sin^2xcos^2 theta = 1

Answer 1

$\bold{\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1}$

Given:

$\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1$

To prove:

$\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1$

Proof:  

LHS

$\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta$

As we see that $\sin ^{6} \theta+\cos ^{6} \theta$  is given so converting it into $\left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}$

Using the value of $\left(a^{3}+b^{3}\right)$ we get the value as $\left(a^{3}+b^{3}\right)=a^{3}+b^{3}+3 a b(a+b)$

Using the value of $\sin ^{2} \theta=a, \cos ^{2} \theta=b$

We get the solution as $\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{3}$

Converting the value of $\sin ^{2} \theta+\cos ^{2} \theta=1$, we get  

$\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{3}=(1)^{3}=1=R H S$

$\bold{\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1}$

Therefore $LHS=RHS$

Hence. Proved.  

Answer 2

A pic is attached with the answer. Hope it helps.