Question

If a straight line makes angles $\alpha ,\beta ,\gamma$ with the coordinate axes, then $\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }+\frac{1}{\sec 2\beta }-2{\sin }^2\gamma$ is equal to

• A. $-1$
• B. $1$
• C. $-2$
• D. $2$
• E. $0$

Answer 1

Answer: (C)

$-2$

If a straight line makes angles $\alpha ,\beta ,\gamma$ with the coordinate axes.
Then, ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$
$\Rightarrow \frac{1+\cos 2\alpha }{2}+\frac{1+\cos 2\beta }{2}+1-{\sin }^2\gamma =1$
$\Rightarrow 1+\cos 2\alpha +1+\cos 2\beta +2-2{\sin }^2\gamma =2$
$\Rightarrow \cos 2\alpha +\cos 2\beta -2{\sin }^2\gamma =-2$
$\Rightarrow \frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }+\frac{1}{\sec 2\beta }-2{\sin }^2\gamma =-2$