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) $\tan \theta + \sec \theta = {e^x}$…..$(i)$

$\therefore \,\,\,\sec \theta - \tan \theta = {e^{ - x}}$…..$(ii)$

From $(i)$ and $(ii),$

$\,2\sec \theta = {e^x} + {e^{ - x}}\,$

$\Rightarrow \,\cos \theta = \frac{2}{{{e^x} + {e^{ - x}}}}.$

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