Question

$\begin{gathered}\begin{gathered} \boxed{\begin{array}{cc}\bf \: if \: \: \: \frac{ {sin}^{4} \alpha }{a} + \frac{ {cos}^{ 4 } \alpha }{b} = \frac{1}{a + b} \: \: \\ \\ \bf \: then \: prove \: that : \\ \\ \\ \bf \: \frac{ {sin}^{4n} \alpha }{ {a}^{2n - 1} } + \frac{ {cos}^{4n} \alpha }{ {b}^{2n - 1} } = \frac{1}{ {(a + b)}^{2n - 1} }\end{array}}\end{gathered}\end{gathered}$​

Answer 1

Step-by-step explanation:

$\color{red} \[ \begin{array}{l} \tt \: Here, \\ \qquad \qquad\tt\dfrac{\sin ^{4} \alpha}{a}+\dfrac{\cos ^{4} \alpha}{b}=\dfrac{1}{a+b} \\ \\ \tt \Rightarrow \quad(a+b)\left(\dfrac{\sin ^{4} \alpha}{a}+\dfrac{\cos ^{4} \alpha}{b}\right)=1 \\ \\ \tt\Rightarrow \quad(a+b)\left(\dfrac{\sin ^{4}\alpha }{a}+\dfrac{\cos ^{4} \alpha}{b}\right)= \left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2} \end{array} \]$

$\color{darkviolet}\begin{array}{l} \tt\Rightarrow \quad \sin ^{4} \alpha+\cos ^{4} \alpha+\frac{b}{a} \sin ^{4} \alpha+\frac{a}{b} \cos ^{4} \alpha =\sin ^{4} \alpha+\cos ^{4} \alpha+2 \sin ^{2} \alpha \cos ^{2} \alpha \\ \\ \tt\Rightarrow \quad \frac{b}{a} \sin ^{4}\alpha +\frac{a}{b} \cos ^{4} \alpha-2 \sin ^{2} \alpha \cos ^{2} \alpha =0 \\ \\ \tt \Rightarrow \quad\left\{\sqrt{\frac{b}{a}} \sin ^{2} \alpha-\sqrt{\frac{a}{b}} \cos ^{2} \alpha\right\}^{2}=0 \\ \\ \tt \Rightarrow \quad \sqrt{\frac{b}{a}} \sin ^{2} \alpha=\sqrt{\frac{a}{b}} \cos ^{2} \alpha \\ \\ \tt \Rightarrow \quad \tan ^{2} \alpha=\frac{a}{b} \end{array}$

$\color{orangered}\[ \begin{array}{l} \tt \Rightarrow \quad \dfrac{\sin ^{2} \alpha}{a}=\dfrac{\cos ^{2} \alpha}{b} \\ \\ \tt\Rightarrow \quad \dfrac{\sin ^{2} \alpha}{a}=\dfrac{\cos ^{2} \alpha}{b}=\dfrac{\sin ^{2} \alpha+\cos ^{2} \alpha}{a+b} \\ \\ \tt \Rightarrow \quad \dfrac{\sin ^{2}\alpha }{a}=\dfrac{\cos ^{2} \alpha}{b}=\dfrac{1}{a+b} \\ \\ \tt \Rightarrow \quad \sin ^{2} \alpha=\dfrac{a}{a+b}, \cos ^{2} \alpha=\dfrac{b}{a+b} \: \qquad \qquad \qquad \ldots(i) \end{array} \]$

$\color{magenta} \[ \begin{array}{l} \text{ Consider,} \\ \\ \tt\dfrac{\sin ^{4 n} \alpha}{a^{2 n-1}}+\dfrac{\cos ^{4 n}\alpha }{b^{2 n-1}}=\dfrac{\left(\sin ^{2} \alpha\right)^{2 n}}{a^{2 n-1}}+ \dfrac{\left(\cos ^{2} \alpha\right)^{2 n}}{b^{2 n-1}} \\ \\ \tt =\dfrac{1}{a^{2 n-1}}\left(\dfrac{a}{a+b}\right)^{2 n}+\dfrac{1}{b^{2 n-1}} \\ \\ \tt\left(\dfrac{b}{a+b}\right)^{2 n} =\dfrac{a}{(a+b)^{2 n}}+\dfrac{b}{(a+b)^{2 n}} \\ \\ = \tt\dfrac{a+b}{(a+b)^{2 n}}=\dfrac{1}{(a+b)^{2 n-1}} \end{array} \]$