Question

If $\frac{sin^{4}\,\theta}{a}+\frac{cos^{4}\,\theta}{b}=\frac{1}{a+b}$, then $\frac{sin^{12}\,\theta}{a^{5}}+\frac{cos^{12}\,\theta}{b^{5}}=$

• A. $1/\left(a+b\right)^{5}$
• B. $1/\left(a-b\right)^{5}$
• C. $1/\left(a^2+b^2\right)^{5}$
• D. $1/\left(a^2+b^2\right)^{3}$

Answer 1

Answer: (A)

Given $\frac{sin^{4}\,\theta}{a}+\frac{cos^{4}\,\theta}{b}=\frac{1}{a+b}$
$\Rightarrow \frac{\left(sin^{2}\,\theta\right)^{2}}{a}+\frac{\left(1-sin^{2}\,\theta\right)^{2}}{b}=\frac{1}{a+b}$
$\Rightarrow \lambda^{2}b\left(a+b\right)+\left(a+b\right)a\left(1-\lambda\right)^{2}=ab$,
where $\lambda=sin^{2}\theta$
$\Rightarrow \lambda^{2}ab+\lambda^{2}b^{2}+a^{2}+a^{2}\lambda^{2}-2a^{2}\lambda+ab+ab\lambda^{2}-2ab\lambda$
$=ab$
$\Rightarrow \lambda^{2}\left\{a^{2}+b^{2}+2ab\right\}+a^{2}-2\lambda a\left(a+b\right)=0$
$\Rightarrow \left\{\lambda\left(a+b\right)\right\}^{2}+a^{2}-2\lambda a\left(a+b\right)=0$
$\Rightarrow \left\{\lambda\left(a+b\right)-a\right\}^{2}=0$
$\Rightarrow \lambda=\frac{a}{a+b}$
$\Rightarrow sin^{2}\theta=\frac{a}{a+b}\quad\ldots\left(i\right)$
From $\left(i\right)$, $cos^{2}\theta=1-sin^{2}\theta=1-\frac{a}{a+b}=\frac{b}{a+b}\quad\ldots\left(ii\right)$
Using $\left(i\right)$ and $\left(ii\right)$, we get
$\frac{sin^{12}\,\theta}{a^{5}}+\frac{cos^{12}\,\theta }{b^{5}}$
$=\frac{\left(sin^{2}\,\theta\right)^{6}}{a^{5}}+\frac{\left(cos^{2}\,\theta\right)^{6}}{b^{5}}$
$=\frac{\left(\frac{a}{a+b}\right)^{6}}{a^{5}}+\frac{\left(\frac{b}{a+b}\right)^{6}}{b^{5}}$
$=\frac{a+b}{\left(a+b\right)^{6}}=\frac{1}{\left(a+b\right)^{5}}$