Question

If $n=\frac{cos\alpha }{cos\beta },m=\frac{sin\alpha }{sin\beta },$ then $(m^2-n^2)si{n}^2\beta$ is

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

Answer 1

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

$(m^2-n^2)si{n}^2\beta =(\frac{si{n}^2\alpha }{si{n}^2\beta }-\frac{co{s}^2\alpha }{co{s}^2\beta })si{n}^2\beta$

$=\frac{si{n}^2\alpha co{s}^2\beta -co{s}^2\alpha si{n}^2\beta }{si{n}^2\beta co{s}^2\beta }si{n}^2\beta$

$=\frac{(1-co{s}^2\alpha )(1-si{n}^2\beta )-co{s}^2\alpha si{n}^2\beta }{{\cos }^2\beta }$

$=\frac{1-co{s}^2\alpha -si{n}^2\beta +co{s}^2\alpha si{n}^2\beta -co{s}^2\alpha si{n}^2\beta }{co{s}^2\beta }$

$=\frac{co{s}^2\beta -co{s}^2\alpha }{co{s}^2\beta }$

$=1-\frac{co{s}^2\alpha }{co{s}^2\beta }$

$=1-n^2$