Write in the simplest form: $\cot^{-1}\left(\sqrt{1+x^{2}}+x\right)$.

$\dfrac{1}{2}\cot^{-1}x\ \left(=\dfrac{\pi}{4}-\dfrac{1}{2}\tan^{-1}x\right)$

CBSE Class 12 · 2007 · Inverse Trigonometric FunctionsLong Answer

Write in the simplest form:

cot^{- 1} (1 + x^{2} + x)

Answer: $\frac{1}{2} cot^{- 1} x (= \frac{π}{4} - \frac{1}{2} tan^{- 1} x)$

Put

x = tan θ

with

θ = tan^{- 1} x \in (- \frac{π}{2}, \frac{π}{2})

, so

1 + x^{2} = sec θ

1 + x^{2} + x = sec θ + tan θ = \frac{1 + sin θ}{cos θ}

Write

1 + sin θ = (cos \frac{θ}{2} + sin \frac{θ}{2})^{2}

and

cos θ = (cos \frac{θ}{2} + sin \frac{θ}{2}) (cos \frac{θ}{2} - sin \frac{θ}{2})

\frac{1 + sin θ}{cos θ} = \frac{cos \frac{θ}{2} + sin \frac{θ}{2}}{cos \frac{θ}{2} - sin \frac{θ}{2}} = \frac{1 + tan \frac{θ}{2}}{1 - tan \frac{θ}{2}} = tan (\frac{π}{4} + \frac{θ}{2}) = cot (\frac{π}{4} - \frac{θ}{2})

Therefore

cot^{- 1} (1 + x^{2} + x) = \frac{π}{4} - \frac{θ}{2} = \frac{π}{4} - \frac{1}{2} tan^{- 1} x = \frac{1}{2} (\frac{π}{2} - tan^{- 1} x) = \frac{1}{2} cot^{- 1} x

\frac{1}{2} cot^{- 1} x

CBSE Class 12 Board · 2007 · Subjective