Trigonometric Functions — MATHS STD 11 Science — Question

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

Solution:

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):

$2\sec\text{x}= \text{k} + \frac{1}{\text{k}}$

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

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

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

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