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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
$\tan \left(60^{\circ}+ A \right) \tan \left(60^{\circ}- A \right)$ is equal to
  • ✓
    $\frac{2 \cos 2 A+1}{2 \cos 2 A-1}$
  • B
    $\frac{2 \cos 2 A-1}{2 \cos 2 A+1}$
  • C
    $\frac{\cos 2 A+21}{\cos 2 A-1}$
  • D
    $\frac{\cos 2 A-1}{\cos 2 A+21}$
✓

Answer

Correct option: A.
$\frac{2 \cos 2 A+1}{2 \cos 2 A-1}$
(A)
$\tan \left(60^{\circ}+ A \right) \tan \left(60^{\circ}- A \right)$
$=\frac{\sin ^2 60^{\circ}-\sin ^2 A}{\cos ^2 60^{\circ}-\sin ^2 A}$
$=\frac{\frac{3}{4}-\left(\frac{1-\cos 2 A}{2}\right)}{\frac{1}{4}-\left(\frac{1-\cos 2 A}{2}\right)}=\frac{3-2+2 \cos 2 A}{1-2+2 \cos 2 A}$
$=\frac{2 \cos 2 A+1}{2 \cos 2 A-1}$

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