MCQ
If $\tan \theta + \sec \theta = {e^x},$ then $\cos \theta $ equals
- A$\frac{{({e^x} + {e^{ - x}})}}{2}$
- ✓$\frac{2}{{({e^x} + {e^{ - x}})}}$
- C$\frac{{({e^x} - {e^{ - x}})}}{2}$
- D$\frac{{({e^x} - {e^{ - x}})}}{{({e^x} + {e^{ - x}})}}$
✓
Answer
Correct option: B.
$\frac{2}{{({e^x} + {e^{ - x}})}}$
b
(b) $\tan \theta + \sec \theta = {e^x}$…..$(i)$
(b) $\tan \theta + \sec \theta = {e^x}$…..$(i)$
$\therefore \,\,\,\sec \theta - \tan \theta = {e^{ - x}}$…..$(ii)$
From $(i)$ and $(ii),$
$\,2\sec \theta = {e^x} + {e^{ - x}}\,$
$\Rightarrow \,\cos \theta = \frac{2}{{{e^x} + {e^{ - x}}}}.$
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