Question

If $si{n}^{-1}\left(x/a\right)+co{s}^{-1}\left(y/b\right)=\alpha$. Then $x^2/a^2+y^2/b^2$ is equal to

• A. $\left(2xy/ab\right)cos\alpha +si{n}^2\alpha$
• B. $\left(2xy/ab\right)sin\alpha +co{s}^2\alpha$
• C. $\left(2xy/ab\right)co{s}^2\alpha +sin\alpha$
• D. $\left(2xy/ab\right)si{n}^2\alpha +cos\alpha$

Answer 1

The correct option is B $\left(2xy/ab\right)sin\alpha +co{s}^2\alpha$

$si{n}^{-1}\left(\frac{x}{a}\right)+co{s}^{-1}\left(\frac{y}{b}\right)=\alpha \Rightarrow si{n}^{-1}\left(\frac{x}{a}\right)+si{n}^{-1}\left(\sqrt{1-{\left(\frac{y}{b}\right)}^2}\right)=\alpha \Rightarrow si{n}^{-1}(\frac{x}{a}\times \frac{y}{b}+\sqrt{1-{\left(\frac{x}{a}\right)}^2}\sqrt{1-{\left(\frac{y}{b}\right)}^2})=\alpha \Rightarrow \frac{x}{a}\times \frac{y}{b}+\sqrt{1-{\left(\frac{x}{a}\right)}^2}\sqrt{1-{\left(\frac{y}{b}\right)}^2}=sin\alpha \Rightarrow \sqrt{1-{\left(\frac{x}{a}\right)}^2}\sqrt{1-{\left(\frac{y}{b}\right)}^2}=sin\alpha -\frac{xy}{ab}\Rightarrow (1-{\left(\frac{x}{a}\right)}^2)(1-{\left(\frac{y}{b}\right)}^2)=si{n}^2\alpha +\frac{x^2{y}^2}{a^2{b}^2}-2sin\alpha \left(\frac{xy}{ab}\right)\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=\left(\frac{2xy}{ab}\right)sin\alpha +co{s}^2\alpha$