Question-70162

Question Number 70162 by TawaTawa last updated on 01/Oct/19

Answered by mind is power last updated on 01/Oct/19

\Rightarrow 1 = (a + b) (\frac{{s i n}^{4} (θ)}{a} + \frac{{c o s}^{4} (θ)}{b}) \dots S

\Rightarrow \frac{b}{a} {S i n}^{4} (θ) + \frac{a}{b} {c o s}^{4} (θ) + {c o s}^{4} (θ) + {s i n}^{4} (θ) = 1

t = \frac{a}{b}

\Rightarrow {t s i n}^{4} (θ) + \frac{{c o s}^{4} (θ)}{t} = 1 - ({c o s}^{4} (θ) + {s i n}^{4} (θ))

\Rightarrow {t s i n}^{4} (θ) + \frac{{c o s}^{4} (θ)}{t} = 2 {c o s}^{2} θ) {s i n}^{2} (θ)

\Rightarrow t \frac{{s i n}^{2} (θ)}{{c o s}^{2} (θ)} + \frac{{c o s}^{2} (θ)}{{t s i n}^{2} (θ)} = 2

\Rightarrow t . {t a n}^{2} (θ) + \frac{1}{t {t a n}^{2} (θ)} = 2

l e t u = t {t a n}^{2} (θ)

\Rightarrow u + \frac{1}{u} = 2 \Rightarrow {(\sqrt{u} - \frac{1}{\sqrt{u}})}^{2} = 0 \Rightarrow u = 1

t . {t a n}^{2} (θ) = 1 \Rightarrow \frac{a}{b} = \frac{{c o s}^{2} (θ)}{{s i n}^{2} (θ)} = \frac{1}{{s i n}^{2} (θ)} - 1

\Rightarrow {S i n}^{2} (θ) = \frac{b}{a + b}

{c o s}^{2} (θ) = \frac{a}{a + b}

{S i n}^{8} (θ) = \frac{b^{4}}{{(a + b)}^{4}}

{c o s}^{8} (θ) = \frac{a^{4}}{{(a + b)}^{3}}

\frac{{S i n}^{8} (θ)}{a^{3}} + \frac{{c o s}^{8} (θ)}{b^{3}} = \frac{b^{4}}{a^{3} {(a + b)}^{4}} + \frac{a^{4}}{b^{3} {(a + b)}^{4}} = \frac{1}{{(a + b)}^{4}} (\frac{b^{4}}{a^{3}} + \frac{a^{4}}{b^{3}}) . . H

{(\frac{b^{3}}{a^{3}} + \frac{a^{3}}{b^{3}})}^{} = {(\frac{b}{a} + \frac{a}{b})}^{3} - 3 (\frac{a}{b} + \frac{b}{a}) = 8 - 6 = 2

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

\frac{b^{4}}{a^{3}} + \frac{a^{4}}{b^{3}} = (a + b) (\frac{b^{3}}{a^{3}} + \frac{a^{3}}{b^{3}} - \frac{b^{2}}{a^{2}} - \frac{a^{2}}{b^{2}} + \frac{b}{a} + \frac{a}{b} - 1) = (a + b) [2 - 2 + 2 - 1) = (a + b)

H = \frac{1}{{(a + b)}^{4}} . (a + b) = \frac{1}{{(a + b)}^{3}}

Commented by TawaTawa last updated on 01/Oct/19

Wow, God bless you sir

Commented by mind is power last updated on 01/Oct/19

y^{'} r e W e l c o m