MCQ
$\sec {50^o} + \tan {50^o}$ is equal to
- A$\tan {20^o} + \tan {50^o}$
- B$2\tan {20^o} + \tan {50^o}$
- ✓$\tan {20^o} + 2\tan {50^o}$
- D$2\tan {20^o} + 2\tan {50^o}$
✓
Answer
Correct option: C.
$\tan {20^o} + 2\tan {50^o}$
c
(c) $\sec {50^o} + \tan {50^o}$
(c) $\sec {50^o} + \tan {50^o}$
==> $\tan ({70^o} - {20^o}) = \frac{{\tan {{70}^o} - \tan {{20}^o}}}{{1 + \tan {{70}^o}\tan {{20}^o}}}$
==> $\tan {50^o} + \tan {70^o}\tan {20^o}\tan {50^o} = \tan {70^o} - \tan {20^o}$
==> $\tan {50^o} + \tan {50^o} = \tan {70^o} - \tan {20^o}$
$[\,\because \tan {70^o} = \cot {20^o}]$
==> $2\tan {50^o} + \tan {20^o} = \tan {70^o}$
==> $2\tan {50^o} + \tan {20^o} = \tan {50^o} + \sec {50^o}$.
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