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