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Introduction of Trigonometry — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsIntroduction of Trigonometry1 Mark
MCQ
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$ =
  • A
    $\cos 60^{\circ}$
  • B
    $\sin 60^{\circ}$
  • C
    $\tan 60^{\circ}$
  • D
    $\sin 30^{\circ}$
✓

Answer

We have $\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$= $\frac{2\times\frac{1} {\sqrt{3}}}{1- (\frac{1} {\sqrt{3}})^{2} }$
= ${\frac2{\sqrt{3}}}\times{}{\frac3{{2}}}$
= ${\frac3{{\sqrt{3}}}}= \sqrt{3} $ = $\tan 60^{\circ}$

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