Question

If $x=aco{s}^4\theta$ and $y=bsi{n}^4\theta$, then $\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=$

• A. $\frac{1}{ab}$
• B. $a^2{b}^2$
• C. $1$
• D. $-1$

Answer 1

The correct option is C $1$

$x=a\times {\cos }^4\theta$

$y=b\times {\sin }^4\theta$

find $\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}},$

$\frac{x}{a}={\cos }^4\theta$

$\frac{y}{b}={\sin }^4\theta$

$\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=\sqrt{{\cos }^4\theta }+\sqrt{{\sin }^4\theta }$

${\cos }^2\theta +{\sin }^2\theta$

$=1$