Skip to content
Miscellaneous Exercise 3 · Q121

Q.Show that tan⁡−11−x1+x=π4−12cos⁡−1x\tan^{-1}\sqrt{\dfrac{1-x}{1+x}} = \dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1}x, for −1≤x≤1-1 \le x \le 1.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
68% · 121/177 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Let x=cos⁡θx=\cos\theta with θ=cos⁡−1x∈[0,π]\theta=\cos^{-1}x\in[0,\pi]. Then 1−x1+x=1−cos⁡θ1+cos⁡θ=tan⁡2θ2\dfrac{1-x}{1+x}=\dfrac{1-\cos\theta}{1+\cos\theta}=\tan^2\dfrac{\theta}{2} (half-angle identity), so 1−x1+x=tan⁡θ2\sqrt{\dfrac{1-x}{1+x}}=\tan\dfrac{\theta}{2} (positive, since θ∈[0,π]  ⟹  θ/2∈[0,π2]\theta\in[0,\pi]\implies\theta/2\in\left[0,\dfrac{\pi}{2}\right], where tangent is non-negative). Since θ/2∈[0,π2)⊂(−π2,π2)\theta/2\in\left[0,\dfrac{\pi}{2}\right)\subset\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right) (excluding the single boundary case θ=π\theta=\pi), $\ …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.