Question

$2({\sin }^{6}A+{\cos }^{6}A)-3({\sin }^{4}A+{\cos }^{4}A)+1=0.$
Another form :
The expression
$3[{\sin }^4(\frac{3}{2}\pi -\alpha )+{\sin }^4(3\pi +4)]-2[{\sin }^6(\frac{1}{2}\pi +\alpha )+{\sin }^6(5\pi -\alpha )]$ is equal to

• A. $0$
• B. $1$
• C. $3$
• D. $\sin 4\alpha +\sin 6\alpha$
• E. none of these

Answer 1

The correct option is A $0$

We know that ${\sin }^2\theta +{\cos }^2\theta =1$ and
$a^3+b^3=(a+b{)}^3-3ab(a+b)$,
$a^2+b^2=(a+b{)}^2-2ab$,
$\therefore L.H.S.=2\{{({\sin }^{2}A+{\cos }^{2}A)}^3-3{\sin }^{2}A{\cos }^{2}A({\sin }^{2}A+{\cos }^{2}A)\}$
$-3\{({\sin }^{2}A+{\cos }^{2}A)-{\sin }^{2}A{\cos }^{2}A\}+1$
$=2-6{\sin }^{2}A{\cos }^{2}A-3+6{\sin }^{2}A{\cos }^{2}A+1$
$=3-3=0$.
Another form :
Ans. $\left(b\right)$. By part $\left(a\right)$.
$L.H.S.=3[{\cos }^4\alpha +{\sin }^4\alpha ]-2[{\cos }^6\alpha +{\sin }^6\alpha ]$