Question

The value of $2({\sin }^6\theta +{\cos }^6\theta )-3({\sin }^4\theta +{\cos }^4\theta )+1$ is

• A. $2$
• B. $0$
• C. $4$
• D. $6$

Answer 1

Answer: (B)

$0$

We have,

$2({\sin }^6\theta +{\cos }^6\theta )-3({\sin }^4\theta +{\cos }^4\theta )+1$

$=2[{\left({\sin }^2\theta \right)}^3+{\left({\cos }^2\theta \right)}^3]-3({\sin }^4\theta +{\cos }^4\theta )+1$

$=2\left[({\sin }^2\theta +{\cos }^2\theta )({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta )\right]-3({\sin }^4\theta +{\cos }^4\theta )+1$

$=2[{\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta ]-3({\sin }^4\theta +{\cos }^4\theta )+1$

$=2{\sin }^4\theta +2{\cos }^4\theta -2{\sin }^2\theta {\cos }^2\theta -3{\sin }^4\theta -3{\cos }^4\theta +1$

$=-{\sin }^4\theta -{\cos }^4\theta -2{\sin }^2\theta {\cos }^2\theta +1$

$=-({\sin }^4\theta +{\cos }^4\theta +2{\sin }^2\theta {\cos }^2\theta )+1$

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

$=-1+1$

$=0$

Hence, this is the answer.