Skip to content
Question 63 of 71

Q.If sin⁡−1x+cot⁡−1(12)=π2\sin^{-1}x+\cot^{-1}\left(\dfrac12\right)=\dfrac{\pi}{2}, then x is equal to :

(a) 25\dfrac{2}{\sqrt5}
(b) 12\dfrac12
(c) 32\dfrac{\sqrt3}{2}
(d) 15\dfrac{1}{\sqrt5}
Puducherry TnboardTamil Nadu HSC (DGE) Board 2024MCQ· 1mImportance★★★★★
89% · 63/71 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 →

Uses sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x+\cos^{-1}x=\pi/2 together with a right-triangle reading of cot⁡−1(1/2)\cot^{-1}(1/2) as cos⁡−1\cos^{-1} of something.

  1. Since sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\dfrac\pi2 for all x∈[−1,1]x\in[-1,1], the given equation sin⁡−1x+cot⁡−1(12)=π2\sin^{-1}x+\cot^{-1}\left(\dfrac12\right)=\dfrac\pi2 means cot⁡−1(12)=cos⁡−1x\cot^{-1}\left(\dfrac12\right)=\cos^{-1}x.
  2. Let θ=cot⁡−1(12)\theta=\cot^{-1}\left(\dfrac12\right), so cot⁡θ=12\cot\theta=\dfrac12, i.e. tan⁡θ=2\tan\theta=2. In a right triangle, opposite =2=2, adjacent =1=1, hypotenuse =1+4=5=\sqrt{1+4}=\sqrt5. …

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.