Trigonometric Functions — MATHS STD 11 Science — Question

2. $\text{b}$

Solution:

Givan: $\tan\text{x}=\frac{​​\text{a}}{\text{b}}$

Now,

$=\cos2\text{x}+\alpha\sin2\text{x}$

$=\text{b}\Big(\frac{1-\tan^2\text{x}}{1+\tan^2\text{x}}\Big)+\text{a}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)$

$=\text{b}\Bigg(\frac{1-\frac{\text{a}^2}{\text{b}^2}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg)+\text{a}\Bigg(\frac{1\times\frac{\text{a}}{\text{b}}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg)$

$=\frac{\text{b}(\text{b}^2-\text{a}^2)}{\text{a}^2+\text{b}^2}+\frac{2\text{a}^2\text{b}^2}{\text{a}^2+\text{b}^2}$

$=\frac{\text{b}^3-\text{a}^2\text{b}+2\text{a}^2\text{b}}{\text{a}^2+\text{b}^2}$

$=\frac{\text{b}^3+\text{a}^2\text{b}}{\text{a}^2+\text{b}^2}$

$=\frac{\text{b}(\text{b}^2+\text{a}^2)}{\text{a}^2+\text{b}^2}$

$=\text{b}$