Question

If $\sin x\cos hy=\cos \theta$ and $\cos x\sin hy=\sin \theta$ then $\sin h^{2}y=?$

• A. ${\sin }^{2}x$
• B. $\cos h^{2}x$
• C. ${\cos }^{2}x$
• D. $1$

Answer 1

The correct option is C

${\cos }^{2}x$

Explanation for the correct option

Square both the sides of the expressions $\sin x\cos hy=\cos \theta$ and $\cos x\sin hy=\sin \theta$, we will get two equations

${\left(\sin x\cos hy\right)}^2={\cos }^2\theta \dots \left(1\right)$

${\left(\cos x\sin hy\right)}^2={\sin }^2\theta \dots \left(2\right)$

Now, add bot the equations,

$$\begin{gathered} \therefore {\left(\sin x\cos hy\right)}^2+{\left(\cos x\sin hy\right)}^2={\cos }^2\theta +{\sin }^2\theta \\ \Rightarrow {\sin }^{2}x\cos h^{2}y+{\cos }^{2}x\sin h^{2}y=1\left[\because {\cos }^2\theta +{\sin }^2\theta =1\right] \end{gathered}$$

Apply the identity $\mathbf{cos}{\mathit{h}}^{\mathbf{2}}\mathit{y}\mathbf{-}\mathbf{sin}{\mathit{h}}^{\mathbf{2}}\mathit{y}\mathbf{=}\mathbf{1}$,

$\begin{array}{rcl}\therefore {\sin }^{2}x\left(1+\sin h^{2}y\right)+{\cos }^{2}x\sin h^{2}y& =& 1\\ & \Rightarrow & {\sin }^{2}x+{\sin }^{2}x\sin h^{2}y+{\cos }^{2}x\sin h^{2}y=1\\ & \Rightarrow & \sin h^{2}y\left({\sin }^{2}x+{\cos }^{2}x\right)=1-{\sin }^{2}x\\ & \Rightarrow & \sin h^{2}y={\cos }^{2}x\end{array}$

Hence, the correct option is (C).