Question

If $\frac{{\cos }^2\theta }{a}=\frac{{\sin }^2\theta }{b}$ then $\frac{{\cos }^4\theta }{a}+\frac{{\sin }^4\theta }{b}=$

• A. $\frac{1}{a+b}$
• B. $\frac{1}{(a+b{)}^2}$
• C. $\frac{1}{a^2}+\frac{1}{b^2}$
• D. $a+b$

Answer 1

Answer: (A)

$\frac{1}{a+b}$

from the given information we conclude that ${\tan }^2\theta =\frac{b}{a}$

so $\tan \theta =\sqrt{\frac{b}{a}}$

so $\sin \theta =\sqrt{\frac{b}{a+b}}$

and $\cos \theta =\sqrt{\frac{a}{a+b}}$

$\Rightarrow \frac{{\cos }^4\theta }{a}+\frac{{\sin }^4\theta }{b}=\frac{a}{(a+b{)}^2}+\frac{b}{(a+b{)}^2}=\frac{a+b}{(a+b{)}^2}=\frac{1}{a+b}$