Trigonometric Functions — MATHS STD 11 Science — Question

2. b

Solution:

Given that, $\tan\theta=\frac{\text{a}}{\text{b}}$ 

$\text{b}\cos2\theta+\text{a}\sin2\theta=\text{b}\Big[\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big]+\text{a}\Big[\frac{2\tan\theta}{1+\tan^2\theta}\Big]$

$=\text{b}\Bigg[\frac{1-\frac{\text{a}^2}{\text{b}^2}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg]+\text{a}\Bigg[\frac{\frac{2\text{a}}{\text{b}}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg]$

$=\text{b}\Big[\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}\Big]+\Bigg[\frac{\frac{2\text{a}^2}{\text{b}}}{\frac{\text{b}^2+\text{a}^2}{\text{b}^2}}\Bigg]$

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

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

Hence, the correct option is (b).