First slide

Trigonometric transformations

Question

If $\frac{a x}{\cos θ} + \frac{b y}{\sin θ} = (a^{2} - b^{2})$ , and $\frac{a x \sin θ}{\cos^{2} θ} - \frac{b y \cos θ}{\sin^{2} θ} = 0$ , then ${(a x)}^{2 / 3} + {(b y)}^{2 / 3} =$

Moderate

A

${(a^{2} - b^{2})}^{2 / 3}$
B

${(a^{2} + b^{2})}^{2 / 3}$
C

${(a^{2} + b^{2})}^{3 / 2}$
D

${(a^{2} - b^{2})}^{3 / 2}$

Solution

From the second relation, we get $\frac{\sin^{2} θ \sin θ}{\cos^{2} θ \cos θ} = \frac{b y}{a x}; \tan^{3} θ = b y / a x$
$∴ \tan θ = {(b y)}^{1 / 3} / {(a x)}^{1 / 3}$
$∴ \sin θ = \frac{{(b y)}^{1 / 3}}{{[{(a x)}^{2 / 3} + {(b y)}^{2 / 3}]}^{1 / 2}}$
$\cos θ = \frac{{(a x)}^{1 / 3}}{{[{(a x)}^{2 / 3} + {(b y)}^{2 / 3}]}^{1 / 2}}$
Putting for $\sin θ$ and $\cos θ$
in $\frac{a x}{\cos θ} + \frac{b y}{\sin θ} = a^{2} - b^{2}$

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