MCQ
${\cos ^2}{76^o} + {\cos ^2}{16^o} - \cos {76^o}\cos {16^o} = $
- A$-1/4$
- B$1/2$
- C$0$
- ✓$3/4$
$ = \frac{1}{2}\left[ {1 + \cos {{152}^o} + 1 + \cos {{32}^o} - \cos {{92}^o} - \cos {{60}^o}} \right]$
$ = \frac{1}{2}\left[ {2 - \frac{1}{2} + \cos {{152}^o} + \cos {{32}^o} - \cos {{92}^o}} \right]$
$ = \frac{1}{2}\left[ {\frac{3}{2} + 2\cos {{92}^o}\cos {{60}^o} - \cos {{92}^o}} \right]$
$ = \frac{1}{2}\left[ {\frac{3}{2} + \cos {{92}^o} - \cos {{92}^o}} \right]$
$ = \frac{3}{4}$.
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$tx -2y-3t=0$
$x - 2ty+ 3 = 0$ $\left( {t \in R} \right)$, is