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 $e^{\sin x}-e^{-\sin x}-4=0,$ then the number of real values of $x$ is

Evaluate the expression: $\sum \frac{1}{1 + x^{a-b} + x^{a-c}}$

If $\frac{1}{\sin 45^{\circ} \sin 46^{\circ}}+\frac{1}{\sin 46^{\circ} \sin 47^{\circ}}+\ldots$ up to $45$ terms $=\frac{1}{\sin x^{\circ}}$,then $\sin \left(\frac{\pi}{2} x\right)=$

$\frac{\cos 13^{\circ}-\sin 13^{\circ}}{\cos 13^{\circ}+\sin 13^{\circ}}+\frac{1}{\cot 148^{\circ}}$ is equal to

Evaluate the sum: $\sum \frac{1}{1 + x^{a - b} + x^{a - c}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module