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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
If $\tan\theta=\frac{\text{a}}{\text{b}},$ then $\text{b}\cos2\theta+\text{a}\sin2\theta$ is equal to:
  • A
    $a$
  • ✓
    $b$
  • C
    $\frac{\text{a}}{\text{b}}$
  • D
    None
✓

Answer

Correct option: B.
$b$
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 $(b).$

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