Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

$\sin x\sin hy=\cos \theta$

Squaring both sides,

${\sin }^{2}x\sin h^{2}y={\cos }^2\theta$ ....................... (1)

And,

$\cos x\cos hy=\sin \theta$

Squaring both sides,

${\cos }^{2}x\cos h^{2}y={\sin }^2\theta$ ....................... (2)

Adding (1) and (2), we get,

${\sin }^{2}x\sin h^{2}y+{\cos }^{2}x\cos h^{2}y={\cos }^2\theta +{\sin }^2\theta$

${\sin }^{2}x\sin h^{2}y+(1-{\sin }^{2}x)\cos h^{2}y=1$

${\sin }^{2}x\sin h^{2}y+\cos h^{2}y-{\sin }^{2}x\cos h^{2}y=1$

${\sin }^{2}x(\sin h^{2}y-\cos h^{2}y)+\cos h^{2}y=1$

$-{\sin }^{2}x+\cos h^{2}y=1$ $(\cos h^{2}y-\sin h^{2}y=1)$

Add $1$ both sides,

$1-{\sin }^{2}x+\cos h^{2}y=1+1$

${\cos }^{2}x+\cos h^{2}y=2$