Question

Let $a$ be maximum value of $(3\cos \theta -4\sin \theta )$ and $\theta \ne \frac{n\pi }{2}$. If $\alpha =a{\sin }^2\theta \cdot {\cos }^3\theta$ and $\beta =a{\sin }^3\theta \cdot {\cos }^2\theta$, then $\sqrt{\frac{{\left({\alpha }^2+{\beta }^2\right)}^5}{(\alpha \beta {)}^4}}=$

• A. $5\sin \frac{\theta }{2}{\cos }^2\frac{\theta }{2}$
• B. $-3\sin \theta$
• C. $5$
• D. $16$

Answer 1

Answer: (C)

Maximum value of $3\cos \theta -4\sin \theta$
$\therefore a=\sqrt{3^2+(-4{)}^2}=5$
$\alpha =5{\sin }^2\theta {\cos }^3\theta$
$\beta =5{\sin }^3\theta {\cos }^2\theta$
Now, ${\alpha }^2+{\beta }^2=5^2\left({\sin }^4\theta {\cos }^6\theta +{\sin }^6\theta {\cos }^4\theta \right)$
$=25\left({\cos }^2\theta +{\sin }^2\theta \right){\sin }^4\theta {\cos }^4\theta$
$=25{\sin }^4\theta {\cos }^4\theta$
$\therefore {\left({\alpha }^2+{\beta }^2\right)}^5={\left(25{\sin }^4\theta {\cos }^4\theta \right)}^5$
$\Rightarrow {\left({\alpha }^2+{\beta }^2\right)}^{5/2}={\left(5{\sin }^2\theta {\cos }^2\theta \right)}^5$
$\alpha \beta =25{\sin }^5\theta {\cos }^5\theta$
$\Rightarrow \sqrt{\frac{{\left({\alpha }^2+{\beta }^2\right)}^5}{(\alpha \beta {)}^4}}=\frac{{\left(5{\sin }^2\theta {\cos }^2\theta \right)}^5}{{\left(5^2{\sin }^5\theta {\cos }^5\theta \right)}^2}$
$=\frac{5^5{\sin }^{10}\theta {\cos }^{10}\theta }{5^4{\sin }^{10}\theta {\cos }^{10}\theta }=5$