If-a-sin-2-x-b-cos-2-x-c-b-sin-2-y-a-cos-2-y-d-and-a-tan-x-b-tan-y-then-a-2-b-2-in-terms-of-a-b-c-d-

Question Number 24973 by Rishabh#1 last updated on 30/Nov/17

If a \sin^{2} x + b \cos^{2} x = c, b \sin^{2} y + a \cos^{2} y = d

and a \tan x = b \tan y then \frac{a^{2}}{b^{2}} = ?

(in terms of a, b, c, d)

Commented by nnnavendu last updated on 30/Nov/17

ans

your question is incomplete

please resend question

{bsin}^{2} y + {acos}^{2} y = ?

Commented by Rishabh#1 last updated on 30/Nov/17

I have updated the question . Thank you

Answered by nnnavendu last updated on 30/Nov/17

ans

{asin}^{2} x + {bcos}^{2} x = c

a (1 - \cos^{2} x) + {bcos}^{2} x = c

a - {acos}^{2} x + {bbcos}^{2} x = c

(a - c) = \cos^{2} x (a - b)

\cos^{2} x = \frac{a - c}{a - b}

\sin^{2} x = 1 - \cos^{2} x

= 1 - (\frac{a - c}{a - b})

= \frac{a - b - a + c}{a - b}

= \frac{c - b}{a - b}

now

{bsin}^{2} y + {acos}^{2} y = d

b (1 - \cos^{2} y) + {acos}^{2} y = d

b - {bcos}^{2} y + {acos}^{2} y = d

(b - d) = \cos^{2} y (b - a)

\cos^{2} y = \frac{b - d}{b - a}

\sin^{2} y = 1 - \cos^{2} y

= 1 - (\frac{b - d}{b - a})

= \frac{b - a - b + d}{b - a}

= \frac{d - a}{b - a}

now

a \tan x = b \tan y

{(\frac{a}{b})}^{2} = {(\frac{tany}{tanx})}^{2}

\frac{a^{2}}{b^{2}} = \frac{\sin^{2} y}{\cos^{2} y} \times \frac{\cos^{2} x}{\sin^{2} x}

= (\frac{\frac{d - a}{b - a}}{\frac{b - d}{b - a}}) \times (\frac{\frac{a - c}{a - b}}{\frac{c - b}{a - b}})

= (\frac{d - a}{b - d}) \times (\frac{a - c}{c - b})

by natural navendu

Commented by Rishabh#1 last updated on 30/Nov/17

t h a n k s a l o t .