Question

If $\frac{si{n}^4\theta }{a}+\frac{co{s}^4\theta }{b}=\frac{1}{a+b}$, then $\frac{si{n}^{12}\theta }{a^5}+\frac{co{s}^{12}\theta }{b^5}=$

• 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 $\frac{si{n}^4\theta }{a}+\frac{co{s}^4\theta }{b}=\frac{1}{a+b}$
$\Rightarrow \frac{{\left(si{n}^2\theta \right)}^2}{a}+\frac{{\left(1-si{n}^2\theta \right)}^2}{b}=\frac{1}{a+b}$
$\Rightarrow {\lambda }^{2}b\left(a+b\right)+\left(a+b\right)a{\left(1-\lambda \right)}^2=ab$,
where $\lambda =si{n}^2\theta$
$\Rightarrow {\lambda }^{2}ab+{\lambda }^2{b}^2+a^2+a^2{\lambda }^2-2a^2\lambda +ab+ab{\lambda }^2-2ab\lambda$
$=ab$
$\Rightarrow {\lambda }^2\left\{a^2+b^2+2ab\right\}+a^2-2\lambda a\left(a+b\right)=0$
$\Rightarrow {\left\{\lambda \left(a+b\right)\right\}}^2+a^2-2\lambda a\left(a+b\right)=0$
$\Rightarrow {\left\{\lambda \left(a+b\right)-a\right\}}^2=0$
$\Rightarrow \lambda =\frac{a}{a+b}$
$\Rightarrow si{n}^2\theta =\frac{a}{a+b}\dots \left(i\right)$
From $\left(i\right)$, $co{s}^2\theta =1-si{n}^2\theta =1-\frac{a}{a+b}=\frac{b}{a+b}\dots \left(ii\right)$
Using $\left(i\right)$ and $\left(ii\right)$, we get
$\frac{si{n}^{12}\theta }{a^5}+\frac{co{s}^{12}\theta }{b^5}$
$=\frac{{\left(si{n}^2\theta \right)}^6}{a^5}+\frac{{\left(co{s}^2\theta \right)}^6}{b^5}$
$=\frac{{\left(\frac{a}{a+b}\right)}^6}{a^5}+\frac{{\left(\frac{b}{a+b}\right)}^6}{b^5}$
$=\frac{a+b}{{\left(a+b\right)}^6}=\frac{1}{{\left(a+b\right)}^5}$