Question

If ${\cos }^{-1}\left(\frac{x}{a}\right)+{\cos }^{-1}\left(\frac{y}{b}\right)=\alpha$, then $\left(\frac{x^2}{a^2}\right)-\left(\frac{2xy}{ab}\right)\cos \alpha +\left(\frac{y^2}{b^2}\right)=$

• A. ${\sin }^2\alpha$
• B. ${\cos }^2\alpha$
• C. ${\tan }^2\alpha$
• D. $co{t}^2\alpha$

Answer 1

The correct option is A

${\sin }^2\alpha$

Explanation for the correct option:

Step 1: Apply inverse trigonometric identity.

Given, ${\cos }^{-1}\left(\frac{x}{a}\right)+{\cos }^{-1}\left(\frac{y}{b}\right)=\alpha$,

By using identity, $\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathit{y}\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{\{}\mathit{x}\mathit{y}\mathbf{-}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\mathbf{.}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\}}$, we get

$\begin{array}{rcl}{\cos }^{-1}\left(\frac{x}{a}\right)+{\cos }^{-1}\left(\frac{y}{b}\right)& =& \alpha \end{array}$

$\Rightarrow$$\begin{array}{rcl}{\cos }^{-1}\{\frac{xy}{ab}-\sqrt{1-\frac{x^2}{a^2}}.\sqrt{1-\frac{y^2}{b^2}}\}& =& \alpha \end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{xy}{ab}-\sqrt{1-\frac{x^2}{a^2}}.\sqrt{1-\frac{y^2}{b^2}}& =& \cos \alpha \end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{xy}{ab}-\cos \alpha & =& \sqrt{1-\frac{x^2}{a^2}}.\sqrt{1-\frac{y^2}{b^2}}\end{array}$

Step 2: take squares on both the sides.

$\begin{array}{rcl}{(\frac{xy}{ab}-\cos \alpha )}^2& =& (1-\frac{x^2}{a^2}).(1-\frac{y^2}{b^2})\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{x^2{y}^2}{a^2{b}^2}-2\frac{xy}{ab}\cos \alpha +{\cos }^2\alpha & =& 1-\frac{y^2}{b^2}-\frac{x^2}{a^2}+\frac{x^2{y}^2}{a^2{b}^2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{x^2}{a^2}+\frac{y^2}{b^2}-2\frac{xy}{ab}\cos \alpha & =& 1-{\cos }^2\alpha \end{array}$

$\begin{array}{rcl}\mathbf{\therefore }\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{2}\mathbf{x}\mathbf{y}}{\mathbf{a}\mathbf{b}}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }& \mathbf{=}& \mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\alpha }\end{array}$

Hence, the correct option is $\left(A\right)$.