Question

${\tan }^2\alpha -{\tan }^2\beta -\left(\frac{1}{2}\right)\sin (\alpha -\beta ){\sec }^2\alpha {\sec }^2\beta$ is zero if

• A. $\sin (\alpha +\beta )=0$
• B. $\sin (\alpha +\beta )=\frac{1}{2}$
• C. $\sin (\alpha -\beta )=0$
• D. $\sin (\alpha -\beta )=\frac{1}{2}$

Answer 1

Answer: (B), (C)

The correct options are
B $\sin (\alpha -\beta )=0$
C $\sin (\alpha +\beta )=\frac{1}{2}$

Let $k={\tan }^2\alpha -{\tan }^2\beta -\frac{1}{2}\sin (\alpha -\beta )se{c}^2\alpha se{c}^2\beta$

$={\tan }^2\alpha -{\tan }^2\beta -\frac{1}{2}\left[\frac{\sin \alpha \cos \beta -\sin \beta \cos \alpha }{co{s}^2\alpha co{s}^2\beta }\right]$

$=\frac{2[{\sin }^2\alpha {\cos }^2\beta -{\sin }^2\beta {\cos }^2\alpha ]}{2{\cos }^2\alpha {\cos }^2\beta }-\frac{[\sin \alpha \cos \beta -\sin \beta \cos \alpha ]}{2{\cos }^2\alpha {\cos }^2\beta }$

$=\frac{(\sin \alpha \cos \beta -\sin \beta \cos \alpha )}{2{\cos }^2\alpha {\cos }^2\beta }[2\sin \alpha \cos \beta +2\sin \beta \cos \alpha -1]$

$=\frac{\sin (\alpha -\beta )}{2{\cos }^2\alpha {\cos }^2\beta }\left[2\right(\sin (\alpha +\beta ))-1]$

If $\sin (\alpha -\beta )=0,$ or $\sin (\alpha +\beta )=\frac{1}{2}$ ,then $k=0$