Question

A line makes angles $\alpha ,\beta ,\gamma$ with the coordinate axes. If $\alpha +\beta =\frac{\pi }{2}$ then ${(\cos \alpha +\cos \beta +\cos \gamma )}^2$ is equal to

• A. $1+\sin 2\alpha$
• B. $1+\cos 2\alpha$
• C. $1-\sin 2\alpha$
• D. $1$

Answer 1

The correct option is A $1+\sin 2\alpha$

as $\alpha ,\beta ,\gamma$ are co-sine angles thus ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1-\left(1\right)\alpha +\beta =\pi /2$

$\alpha =(\pi /2+\beta ),$ Pitting value $y\alpha$ in $\left(1\right)$

${\cos }^2\beta +{\cos }^2(\pi /2-\beta )+{\cos }^2\gamma =1$

${\cos }^2\beta +{\sin }^2\beta +{\cos }^2\gamma =1$

${\cos }^2\gamma =0\Rightarrow \cos \gamma =0$

Let $L=(\cos \alpha +\cos \beta +\cos \gamma {)}^2$

$L=(\cos \alpha +\cos \beta {)}^2$

$L={\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta$

$={\cos }^2\alpha +{\cos }^2(\pi /2-\alpha )+2\cos \alpha \cos (\pi /2-\alpha )$

$={\cos }^2\alpha +{\sin }^2\alpha +2\cos \alpha \sin \alpha [{\sin }^2\theta +{\cos }^2\theta =1\cos (\pi /2=\theta =\sin \theta )$

$L=2+\sin 2x$