Question

If $\alpha ,\beta ,\gamma ,\delta$ be four angles of a cyclic quadrilateral taken in clockwise direction then the value of $(2+\sum cos\alpha cos\beta )$ will be :

• A. $si{n}^2\alpha +si{n}^2\beta$
• B. $co{s}^2\gamma +co{s}^2\delta$
• C. $si{n}^2\alpha +si{n}^2\delta$
• D. $co{s}^2\beta +co{s}^2\gamma$

Answer 1

The correct option is C $si{n}^2\alpha +si{n}^2\delta$

We know that if ABCD be a cyclic quadrilateral then $\alpha +\gamma =\pi$ and $\beta +\delta =\pi$
or $cos\alpha =cos(\pi -\gamma )=cos\gamma$
$\therefore cos\alpha +cos\gamma =0$ ...(1)
and $cos\beta +cos\delta =0$ ....(2)
$\therefore cos\alpha +cos\beta +cos\gamma +cos\delta =0$
Square
$co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma +co{s}^2\delta +2\sum cos\alpha cos\beta =0$
or $2(co{s}^2\alpha +co{s}^2\delta )+2\sum cos\alpha cos\beta =0$ by (1 , 2)
$(1-si{n}^2\alpha )+(1-si{n}^2\delta )+\sum cos\alpha cos\beta =0$
$\therefore 2+\sum cos\alpha cos\beta =si{n}^2\alpha +si{n}^2\delta \Rightarrow \left(c\right)$