Question

If $\alpha co{s}^{2}3\theta +\beta co{s}^4\theta =16co{s}^6\theta +9co{s}^2\theta$ is an identity then-

• A. $\alpha =1,\beta =18$
• B. $\alpha =1,\beta =24$
• C. $\alpha =3,\beta =24$
• D. $\alpha =4,\beta =2$

Answer 1

The correct option is B $\alpha =1,\beta =24$

$\alpha {\cos }^{2}3\theta +\beta {\cos }^4\theta =16{\cos }^6\theta +9{\cos }^2\theta$

$\Rightarrow$ $\alpha (4{\cos }^3\theta -3\cos \theta {)}^2+\beta {\cos }^4\theta =16{\cos }^6\theta +9{\cos }^2\theta$

$\Rightarrow$ $\alpha (16{\cos }^6\theta +9{\cos }^2\theta -24{\cos }^4\theta )+\beta {\cos }^4\theta =16{\cos }^6\theta +9{\cos }^2\theta$

$\Rightarrow$ $\alpha (16{\cos }^6\theta +9{\cos }^2\theta )-24\alpha {\cos }^4\theta +\beta {\cos }^4\theta =(16{\cos }^6\theta +9{\cos }^2\theta )$

$\Rightarrow$ $\alpha (16{\cos }^6\theta +9{\cos }^2\theta )-{\cos }^4\theta (24\alpha -\beta )=(16{\cos }^6\theta +9{\cos }^2\theta )$

$\Rightarrow$ $\alpha (16{\cos }^6\theta +9{\cos }^2\theta )-{\cos }^4\theta (24\alpha -\beta )=(16{\cos }^6\theta +9{\cos }^2\theta )+{\cos }^4\theta \times 0$

Now, we are going to compare,

$\alpha (16{\cos }^6+9{\cos }^2\theta )=16{\cos }^6+9{\cos }^2\theta$

$\therefore$ $\alpha =1$

Again,

$\Rightarrow$ $-{\cos }^4\theta (24\alpha -\beta )={\cos }^4\theta \times 0$

$\Rightarrow$ $24\left(1\right)-\beta =0$ [ Since, $\alpha =1$ ]

$\Rightarrow$ $\beta =24$

$\therefore$ $\alpha =1$ and $\beta =24$