If + =k, = - Vidyadip

MCQ

If $\sec\text{x}+\tan\text{x}=\text{k},\cos\text{x}=$

A
$\frac{\text{x}^2+1}{2\text{k}}$
✓
$\frac{2\text{k}}{\text{x}^2+1}$
C
$\frac{\text{k}}{\text{x}^2+1}$
D
$\frac{\text{k}}{\text{x}^2-1}$

✓

Answer

Correct option: B.

$\frac{2\text{k}}{\text{x}^2+1}$

We have:
$\sec\text{x} +\tan\text{x} = \text{k}\cdots(1)$
$\Rightarrow\frac{1}{\sec\text{x} + \tan\text{x}}=\frac{1}{\text{k}}$
$\Rightarrow\frac{\sec^2\text{x}-\tan^2\text{x}}{\sec\text{x}+\tan\text{x}} = \frac{1}{\text{k}}$
$\Rightarrow\frac{(\sec\text{x} + \tan\text{x})(\sec\text{x}-\tan\text{x})}{(\sec\text{x} + \tan\text{x})} = \frac{1}{\text{k}}$
$\therefore\sec\text{x} - \tan\text{x} = \frac{1}{\text{k}}\cdots(2)$
Adding $(1)$ and $(2):$
$2\sec\text{x}= \text{k} + \frac{1}{\text{k}}$
$\Rightarrow 2\sec\text{x} = \frac{\text{k}^2 + 1}{\text{k}}$
$\Rightarrow \sec\text{x} = \frac{\text{k}^2+1}{2\text{k}}$
$\Rightarrow\frac{1}{\cos\text{x}}= \frac{\text{k}^2 + 1}{2\text{k}}$
$\Rightarrow\cos \text{x} = \frac{2\text{k}}{\text{k}^2 + 1}$

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