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Introduction of Trigonometry — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsIntroduction of Trigonometry1 Mark
MCQ
If $\tan\theta=\frac{\text{a}}{\text{b}},$ then $\frac{\text{a sin}\theta+\text{b cos}\theta}{\text{a sin}\theta-\text{b cos}\theta}$ is equal to :
  • ✓
    $\frac{\text{a}^2+\text{b}^2}{\text{a}^2-\text{b}^2}$
  • B
    $\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$
  • C
    $\frac{\text{a}+\text{b}}{\text{a}-\text{b}}$
  • D
    $\frac{\text{a}-\text{b}}{\text{a}+\text{b}}$
✓

Answer

Correct option: A.
$\frac{\text{a}^2+\text{b}^2}{\text{a}^2-\text{b}^2}$
We have,
$\tan\theta=\frac{\text{a}}{\text{b}}$
In $\triangle \text{ABC}.$

$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\text{AC}^2=\text{a}^2+\text{b}^2$
$\text{AC}=\sqrt{\text{a}^2+\text{b}^2}$
$\therefore \sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2}}$
$\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2}}$
Now, $\frac{\text{a sin}\theta+\text{b cos}\theta}{\text{a sin}\theta-\text{b cos}\theta}$
$=\frac{\text{a}\times\frac{\text{a}}{\sqrt{a^2+b^2}}+\text{b}\times\frac{\text{b}}{\sqrt{a^2+b^2}}}{\text{a}\times\frac{a}{\sqrt{a^2+b^2}}-\text{b}\times\frac{\text{b}}{\sqrt{a^2+^2}}}$
$= \frac{\frac{\text{a}^2+\text{b}^2}{\sqrt{\text{a}^2+\text{b}^2}}}{\frac{\text{a}^2-\text{b}^2}{\sqrt{\text{a}^2+\text{b}^2}}}$
$= \frac{\text{a}^2+\text{b}^2}{\text{a}^2-\text{b}^2}$
Hence the correct option is $(a)$

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