Question

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

• A. $2$
• B. $0$
• C. $4$
• D. $6$

Answer 1

The correct option is B

$0$

Explanation for the correct answer:

Let $S=2\left({\sin }^6\theta +{\cos }^6\theta \right)-3\left({\sin }^4\theta +{\cos }^4\theta \right)+1$

$\Rightarrow$ $S=2\left({\left({\sin }^2\theta \right)}^3+{\left({\cos }^2\theta \right)}^3\right)-3\left({\left({\sin }^2\theta \right)}^2+{\left({\cos }^2\theta \right)}^2\right)+1$ $\left[\because \left(a^3+b^3\right)={\left(a+b\right)}^3-3ab\left(a+b\right),\left(a^2+b^2={\left(a+b\right)}^2-2ab\right)\right]$

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

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

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

$\Rightarrow$ $S=0$

Hence, the value of $2\left({\sin }^6\theta +{\cos }^6\theta \right)-3\left({\sin }^4\theta +{\cos }^4\theta \right)+1$ is $0$.

Hence, option $\left(B\right)$ is the right answer.