Prove that:

Solution:

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

Formula Used:

1) $\cos 2 A=2 \cos ^{2} A-1$

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

Proof:

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

$=\cos ^{-1}\left(2 x^{2}-1\right) \ldots$(1)

Let $x=\cos A \ldots(2)$

Substituting (2) in (1),

$\mathrm{LHS}=\cos ^{-1}\left(2 \cos ^{2} \mathrm{~A}-1\right)$

$=\cos ^{-1}(\cos 2 \mathrm{~A})$

$=2 \mathrm{~A}$

From $(2), A=\cos ^{-1} x$

$2 A=2 \cos ^{-1} x$

$=\mathrm{RHS}$

Therefore, LHS = RHS

Hence proved.