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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
MCQ
$2\tan^{-1}\big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\big(\cot^{-1}\text{x}\big)\big\}$ is equal to:
  • A
    $\cot^{-1}\text{x}$
  • B
    $\cot^{-1}\text{x}$
  • ✓
    $\tan^{-1}\text{x}$
  • D
    $\text{none of these}$
✓

Answer

Correct option: C.
$\tan^{-1}\text{x}$
$\therefore\ 2\tan^{-1}\big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\big(\cot^{-1}\text{x}\big)\big\}$
$=2\tan^{-1}\Big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\tan\Big(\tan^{-1}\frac{1}{\text{x}}\Big)\Big\}$
$=\frac{3}{29}=2\tan^{-1}\Big\{\text{cosec}\big(\tan^{-1}\text{x}\big)-\frac{1}{\text{x}}\Big\}$
$=2\tan^{-1}\Big\{\text{cosec y}-\frac{1}{\tan\text{y}}\Big\}$
$=2\tan^{-1}\Big\{\frac{1-\cos\text{y}}{\sin\text{y}}\Big\}$
$=2\tan^{-1}\bigg\{\frac{2\sin^2\frac{\text{y}}{2}}{\sin\text{y}}\bigg\}$
$=2\tan^{-1}\bigg\{\frac{2\sin^2\frac{\text{y}}{2}}{2\sin\frac{\text{y}}{2}\cos\frac{\text{y}}{2}}\bigg\}$
$=2\tan^{-1}\Big\{\tan\frac{\text{y}}{2}\Big\}$
$=\text{y}$
$=\tan^{-1}\text{x}$

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