Question

If $\frac{\cos x}{\cos y}=n,\frac{\sin x}{\sin y}=m$, then what is $(m^2-n^2){\sin }^{2}y$ equal to?

• A. $1-n^2$
• B. $1+n^2$
• C. $n^2$
• D. $m^2$

Answer 1

The correct option is A $1-n^2$

$co{s}^{2}x=n^{2}co{s}^{2}y$ & $si{n}^{2}x=m^{2}si{n}^{2}y$

$\Rightarrow si{n}^{2}x+co{s}^{2}x=m^{2}si{n}^{2}y+n^2(1-si{n}^{2}y)$

$\Rightarrow 1=(m^2-n^2)si{n}^{2}y+n^2\Rightarrow (1-n^2)=(m^2-n^2)si{n}^{2}y$