If theta is eliminated from the equations x = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given equations are $\frac{x}{a} = \cos(	heta - \alpha)$ and $\frac{y}{b} = \cos(	heta - \beta)$.We know that $(\alpha - \beta) = (	heta - \bet

Vedclass · Accepted Answer

$\sin^2(\alpha - \beta)$

Vedclass · Answer

(B) Given equations are $\frac{x}{a} = \cos(	heta - \alpha)$ and $\frac{y}{b} = \cos(	heta - \beta)$.We know that $(\alpha - \beta) = (	heta - \beta) - (	heta - \alpha)$.Using the cosine difference formula,$\cos(\alpha - \beta) = \cos((	heta - \beta) - (	heta - \alpha)) = \cos(	heta - \beta) \cos(	heta - \alpha) + \sin(	heta - \beta) \sin(	heta - \alpha)$.Substituting the given values,$\cos(\alpha - \beta) = \frac{y}{b} \cdot \frac{x}{a} + \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}}$.Rearranging,$\cos(\alpha - \beta) - \frac{xy}{ab} = \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}}$.Squaring both sides,$\cos^2(\alpha - \beta) + \frac{x^2y^2}{a^2b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = (1 - \frac{x^2}{a^2})(1 - \frac{y^2}{b^2})$.Expanding the right side,$\cos^2(\alpha - \beta) + \frac{x^2y^2}{a^2b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2} + \frac{x^2y^2}{a^2b^2}$.Canceling $\frac{x^2y^2}{a^2b^2}$ from both sides and rearranging,$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = 1 - \cos^2(\alpha - \beta)$.Therefore,the expression is equal to $\sin^2(\alpha - \beta)$.

If $\theta$ is eliminated from the equations $x = a \cos(\theta - \alpha)$ and $y = b \cos(\theta - \beta)$,then $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos(\alpha - \beta)$ is equal to

Similar Questions

If $\sin \theta + \text{cosec} \theta = 2$,the value of $\sin^{10} \theta + \text{cosec}^{10} \theta$ is

If $A$ is in the third quadrant and $\tan A = \frac{\sqrt{7}}{3}$,then $18 - 16 \sin^2 \frac{A}{2} - 32 \sin \frac{A}{2} \sin \frac{5A}{2} = $

If $\operatorname{cosec} \theta + \cot \theta = 5$, then $\sin \theta = $

If $\cos (\theta - \alpha ), \cos \theta$ and $\cos (\theta + \alpha )$ are in $H.P.$, then $\cos \theta \sec \frac{\alpha }{2}$ is equal to:

The value of $\log _{10} \tan 1^{\circ} + \log _{10} \tan 2^{\circ} + \dots + \log _{10} \tan 89^{\circ}$ is equal to :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module