Question

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

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

Answer 1

The correct option is C $\frac{1}{{\left(a+b\right)}^3}$

Let $\frac{{\sin }^4\theta }{a^2}=\frac{{\cos }^4\theta }{b^2}=k$

$⟹{\sin }^4\theta =a^{2}k,{\cos }^4\theta =b^{2}k$

$⟹{\sin }^2\theta =a\sqrt{k},{\cos }^2\theta =b\sqrt{k}$

As we know that

${\sin }^2\theta +{\cos }^2\theta =1⟹a\sqrt{k}+b\sqrt{k}=1⟹k=\frac{1}{(a+b{)}^2}$

$\frac{{\sin }^8}{a^3}+\frac{{\cos }^8\theta }{b^3}=\frac{({\sin }^4\theta {)}^2}{a^3}+\frac{({\cos }^4\theta {)}^2}{b^3}=\frac{a^4{k}^2}{a^3}+\frac{b^4{k}^2}{b^3}=(a+b)\times k^2=(a+b)\times \frac{1}{(a+b{)}^4}=\frac{1}{(a+b{)}^3}$