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

1.	Prove that:\(\sec^{-1}\left(\frac{1}{2\mathrm x^2-1}\right)=2\cos^{-1}\mathrm x\)
Answer» 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.

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

Answer»

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.