{cos ^2}A{(3 - 4{cos ^2}A)^2} + {sin ^2}A{(3 | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c) ${\cos ^2}A{(3 - 4{\cos ^2}A)^2} + {\sin ^2}A{(3 - 4{\sin ^2}A)^2}$ 

$ = {(3\cos A - 4{\cos ^3}A)^2} + {(3\sin A - 4{\sin ^3}A)^2}$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(c) ${\cos ^2}A{(3 - 4{\cos ^2}A)^2} + {\sin ^2}A{(3 - 4{\sin ^2}A)^2}$ 

$ = {(3\cos A - 4{\cos ^3}A)^2} + {(3\sin A - 4{\sin ^3}A)^2}$ 

$ = {(\cos 3A)^2} + {(\sin 3A)^2} = 1$. 

Trick : Put $A = \frac{\pi }{2},{0^o}$, the value of expression remains $1$,

therefore it is independent of $A$ and is equal to $1.$

${\cos ^2}A{(3 - 4{\cos ^2}A)^2} + {\sin ^2}A{(3 - 4{\sin ^2}A)^2} = $

Similar Questions

If $\sin A + \cos A = \sqrt 2 ,$ then ${\cos ^2}A = $

If $A + B + C = {180^o},$ then the value of $(\cot B + \cot C)$ $(\cot C + \cot A)\,\,(\cot A + \cot B)$ will be

$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $

The value of $\frac{{3 + \cot {{76}^o}\cot {{16}^o}}}{{\cot {{76}^o} + \cot {{16}^o}}}$

$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = $