Question-34064

Question Number 34064 by math1967 last updated on 30/Apr/18

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

l e t {s i n}^{2} α = t

\frac{t^{2}}{a} + \frac{{(1 - t)}^{2}}{b} = \frac{1}{a + b}

{b t}^{2} + a (1 - 2 t + t^{2}) = \frac{a b}{a + b}

t^{2} (a + b) + t (- 2 a) + a - \frac{a b}{a + b} = 0

{(a + b)}^{2} t^{2} + t (- 2 a) (a + b) + a^{2} = 0

{(a + b) t - a}^{2} = 0

t = \frac{a}{a + b}

{s i n}^{2} α = \frac{a}{a + b}, {c o s}^{2} α = \frac{b}{a + b}

\frac{{s i n}^{12} α}{a^{5}} + \frac{{c o s}^{12} α}{b^{5}}

\frac{a^{6}}{{(a + b)}^{6} . a^{5}} + \frac{b^{6}}{{(a + b)}^{6} . b^{5}}

\frac{a + b}{{(a + b)}^{6}}

\frac{1}{{(a + b)}^{5}} T a n m a y c h a u d h u r y