Question

If $a^2{\cos }^4\theta -b^2{\sin }^4\theta =0$, then $\frac{{\sin }^8\theta }{a^3}+\frac{{\cos }^8\theta }{b^3}$ is

• A. $\frac{1}{a+b}$
• B. $\frac{1}{(a+b{)}^2}$
• C. $\frac{1}{(a+b{)}^3}$
• D. $\frac{2}{(a+b{)}^4}$

Answer 1

Answer: (C)

$\frac{1}{(a+b{)}^3}$

Given,

$a^2{\cos }^4\theta -b^2{\sin }^4\theta =0$

then ${\tan }^4\theta =\frac{a^2}{b^2}$

${\tan }^2\theta =\frac{a}{b}$

${\sec }^2\theta =\frac{a+b}{b}$

${\cos }^2\theta =\frac{b}{a+b}$

${\sin }^2\theta =\frac{a}{a+b}$

$\frac{{\sin }^8\theta }{a^3}+\frac{{\cos }^8\theta }{b^3}=\frac{a^4}{a^3(a+b{)}^4}+\frac{b^4}{b^3(a+b{)}^4}$

$\frac{a}{(a+b{)}^4}+\frac{b}{(a+b{)}^4}$

$=\frac{1}{(a+b{)}^3}$