Prove that: sec^1(1/2x^2 - 1) = 2cos^-1x

Question

Prove that:

\(\sec^{-1}\left(\frac{1}{2\mathrm x^2-1}\right)=2\cos^{-1}\mathrm x\)

Answer 1

To Prove: \(\sec^{-1}\left(\frac{1}{2\mathrm x^2-1}\right)=2\cos^{-1}\mathrm x\)

Formula Used: 

1) cos 2A = 2 cos 2 A – 1 

2) \(\cos^{-1}A= \sec^{-1}\frac{1}{A}\)

Proof: 

LHS \(=\sec^{-1}\left(\frac{1}{2\mathrm x^2-1}\right)\)

= cos ^-1 (2x ² – 1)… (1) 

Let x = cos A … (2) 

Substituting (2) in (1), 

LHS = cos ^-1 (2 cos ² A – 1) 

= cos ^-1 (cos 2A) 

= 2A 

From (2), A = cos ^-1 x, 

2A = 2 cos ^-1 x 

= RHS 

Therefore, LHS = RHS 

Hence proved.