Question

Please solve 13 (i ).

Answer 1

Answer:

Step-by-step explanation:

Given :

$cos^{-1}\frac{x}{a} + cos^{-1}\frac{y}{b} = \alpha$

Then, prove :

$\frac{x^2}{a^2} - (\frac{2xy}{ab})cos\alpha + \frac{y^2}{b^2} = sin^2\alpha$

Proof :

Let $cos^{-1}\frac{x}{a} = X$

⇒ $cosX=\frac{x}{a}$,

⇒ $sinX = \sqrt{1-cos^2X} = \sqrt{1-(\frac{x}{a})^2 }$

Let $cos^{-1}\frac{y}{b} = Y$

⇒ $cosY=\frac{y}{ab}$.

⇒ $sinY = \sqrt{1-cos^2Y} = \sqrt{1-(\frac{y}{b})^2 }$

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Then,

$cos^{-1}\frac{x}{a} + cos^{-1}\frac{y}{b} = \alpha$

⇒ $X + Y = \alpha$

⇒⇒ $cos \alpha = cos(X+Y) =cosXcosY - sinXsinY$

⇒ $(\frac{x}{a}) (\frac{y}{b}) - [\sqrt{1-(\frac{x}{a})^2 }][\sqrt{1-(\frac{y}{b})^2 } ] = cos \alpha$

⇒ $\frac{xy}{ab} - cos \alpha = [\sqrt{1-(\frac{x}{a})^2 }][\sqrt{1-(\frac{y}{b})^2 } ]$

By squaring both the sides,

We get,

⇒ $(\frac{xy}{ab} - cos \alpha)^2 = ([\sqrt{1-(\frac{x}{a})^2 }][\sqrt{1-(\frac{y}{b})^2 } ])^2$

⇒ $(\frac{xy}{ab})^2 - 2(\frac{xy}{ab})(cos\alpha) + (cos \alpha)^2 = (1-[\frac{x}{a}]^2)(1-[\frac{y}{b}]^2)$

⇒ $\frac{x^2y^2}{a^2b^2} - 2\frac{xy}{ab}cos\alpha + cos^2 \alpha = 1 - \frac{x^2}{a^2} -\frac{y^2}{b^2} + \frac{x^2y^2}{a^2b^2}$

⇒ $\frac{x^2y^2}{a^2b^2} - \frac{x^2y^2}{a^2b^2} + \frac{x^2}{a^2} + \frac{y^2}{b^2}- \frac{2xy}{ab}cos\alpha + cos^2 \alpha = 1$

⇒ $\frac{x^2}{a^2} -\frac{2xy}{ab}cos\alpha+\frac{y^2}{b^2} = 1- cos^2\alpha = sin^2\alpha$

Hence, Proved