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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds1 Mark
MCQ
If $\tan\text{X}=\frac{\text{a}}{\text{b}},$ then $\text{b}\cos2\text{x}+\text{a}\sin2\text{x}$ is equal to:
  • A
    $\text{a}$
  • ✓
    $\text{b}$
  • C
    $\frac{\text{a}}{\text{b}}$
  • D
    $\frac{\text{b}}{\text{a}}$
✓

Answer

Correct option: B.
$\text{b}$
Givan: $\tan\text{x}=\frac{​​\text{a}}{\text{b}}$
Now,
$=\cos2\text{x}+\alpha\sin2\text{x}$
$=\text{b}\Big(\frac{1-\tan^2\text{x}}{1+\tan^2\text{x}}\Big)+\text{a}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)$
$=\text{b}\Bigg(\frac{1-\frac{\text{a}^2}{\text{b}^2}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg)+\text{a}\Bigg(\frac{1\times\frac{\text{a}}{\text{b}}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg)$
$=\frac{\text{b}(\text{b}^2-\text{a}^2)}{\text{a}^2+\text{b}^2}+\frac{2\text{a}^2\text{b}^2}{\text{a}^2+\text{b}^2}$
$=\frac{\text{b}^3-\text{a}^2\text{b}+2\text{a}^2\text{b}}{\text{a}^2+\text{b}^2}$
$=\frac{\text{b}^3+\text{a}^2\text{b}}{\text{a}^2+\text{b}^2}$
$=\frac{\text{b}(\text{b}^2+\text{a}^2)}{\text{a}^2+\text{b}^2}$
$=\text{b}$

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