First slide

Trigonometric Identities

Question

If $a \sin^{2} x + b \cos^{2} x = c, b \sin^{2} y + a \cos^{2} y = d$ and $a \tan x = b \tan y t h e n \frac{a^{2}}{b^{2}} =$

Moderate

A

$\frac{(a - d) (c - a)}{(b - c) (d - b)}$
B

$\frac{(b - c) (b - d)}{(a - c) (a - d)}$
C

$\frac{(b - c) (d - b)}{(a - d) (c - a)}$
D

$\frac{(d - a) (c - a)}{(b - c) (d - b)}$

Solution

$W e h a v e$
$\begin{array}{l} a \sin^{2} x + b \cos^{2} x = c \\ \Rightarrow a \tan^{2} x + b = c \sec^{2} x \\ \Rightarrow (a - c) \tan^{2} x = c - b \Rightarrow \tan^{2} x = \frac{c - b}{a - c} \\ Similarly \tan^{2} y = \frac{d - a}{b - d} \\ b u t a \tan x = b \tan y \Rightarrow \frac{a^{2}}{b^{2}} = \frac{\tan^{2} y}{\tan^{2} x} \\ = \frac{d - a}{b - d} \frac{a - c}{c - b} \\ = \frac{(a - d) (c - a)}{(b - c) (d - b)} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App