Question

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

Answer 1

Answer:

Please substitute 8 by 4n and 3 by 2n-1

Step-by-step explanation:

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