Question

${{\cos }^{-1}}\left( \cos \left( 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right) \right)$ is equal to:
[a] $\sqrt{2}-1$
[b] $\dfrac{\pi }{4}$
[c] $\dfrac{3\pi }{4}$
[d] None of the above.

Answer 1

Hint: Use the fact that $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{2}-1$ and then use the identities $\cot x=\tan \left( \dfrac{\pi }{2}-x \right),{{\cot }^{-1}}\left( \cot x \right)=x+n\pi $ where n is suitably chosen so that the value of $x+n\pi $ is within the interval $\left( 0,\pi \right)$. Alternatively use ${{\cos }^{-1}}\left( \cos x \right)=x+2m\pi $ , where m is suitably chosen so that the value of $x+2m\pi $ is within the interval $\left[ 0,\pi \right]$ and the assume the expression be equal to y. Take cot on both sides and then use $\cot 2x=\dfrac{{{\cot }^{2}}x-1}{2\cot x}$. Simplify to get the expression for coty. Find in the interval $\left( 0,\pi \right)$ the value at which the equation is satisfied. The value of y found is the answer of the question.

Complete step-by-step answer:
We know that $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{2}-1$
Hence, we have $\tan \left( \dfrac{\pi }{2}-\dfrac{3\pi }{8} \right)=\sqrt{2}-1$
We know that $\cot x=\tan \left( \dfrac{\pi }{2}-x \right)$
Put $x=\dfrac{3\pi }{8}$, we have
$\cot \left( \dfrac{3\pi }{8} \right)=\tan \left( \dfrac{\pi }{2}-\dfrac{3\pi }{8} \right)$
Hence, we have
$\cot \left( \dfrac{3\pi }{8} \right)=\sqrt{2}-1$
Hence, we have
${{\cot }^{-1}}\left( \sqrt{2}-1 \right)={{\cot }^{-1}}\left( \cot \left( \dfrac{3\pi }{8} \right) \right)=\dfrac{3\pi }{8}+n\pi $
Since \[\dfrac{3\pi }{8}\in \left( 0,\pi \right)\], we have n = 0.
Hence ${{\cot }^{-1}}\left( \sqrt{2}-1 \right)=\dfrac{3\pi }{8}$
Hence, we have $\cos \left( 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right)=\cos \left( 2\dfrac{3\pi }{8} \right)=\cos \left( \dfrac{3\pi }{4} \right)=\cos \left( \pi -\dfrac{\pi }{4} \right)$
Now, we know that $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ and $\cos \left( \pi -x \right)=-\cos x$
Put $x=\dfrac{\pi }{4}$, we get
$\cos \left( \pi -\dfrac{\pi }{4} \right)=-\cos \left( \dfrac{\pi }{4} \right)=\dfrac{-1}{\sqrt{2}}$
Hence , we have $\cos \left( 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right)=-\dfrac{1}{\sqrt{2}}$
Now, we know that $\arccos \left( -x \right)=\pi -\arccos x$
Hence, we have${{\cos }^{-1}}\left( \cos \left( 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right) \right)={{\cos }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)=\pi -{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)$
Since $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ , we have $\arccos \left( \dfrac{1}{\sqrt{2}} \right)=\dfrac{\pi }{4}$
Hence, we have
${{\cos }^{-1}}\left( \cos \left( 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right) \right)=\pi -\dfrac{\pi }{4}=\dfrac{3\pi }{4}$
Hence, option [c] is correct.

Note: Alternatively, we can use ${{\cos }^{-1}}\left( \cos x \right)=x+2m\pi $ , where m is suitably chosen so that the value of $x+2m\pi $ is within the interval $\left[ 0,\pi \right]$.
Hence, we get
${{\cos }^{-1}}\left( \cos \left( 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right) \right)=2m\pi +2{{\cot }^{-1}}\left( \sqrt{2}-1 \right)$.
Now let $y=2m\pi +2{{\cot }^{-1}}\left( \sqrt{2}-1 \right)$
Taking cot on both sides, we get
$\cot y=\cot \left( 2m\pi +2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right)$
Now since cotx is periodic with period $\pi $, we have
$\cot y=\cot \left( 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right)$
Using $\cot 2x=\dfrac{{{\cot }^{2}}x-1}{2\cot x}$ , we get
$\cot y=\dfrac{{{\cot }^{2}}\left( {{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right)-1}{2\cot \left( {{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right)}$
Now we know that $\cot \left( {{\cot }^{-1}}x \right)=x$
Hence, we have
$\cot y=\dfrac{{{\left( \sqrt{2}-1 \right)}^{2}}-1}{2\left( \sqrt{2}-1 \right)}=\dfrac{2-2\sqrt{2}}{2\left( \sqrt{2}-1 \right)}=-1$
Hence, we have
$y=\dfrac{3\pi }{4}$, which is the same as obtained above.
[2] You can prove that $\tan \left( \dfrac{\pi }{8} \right)=\sqrt{2}-1$, by putting $x=\dfrac{\pi }{4}$ in the identity $\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$.
Hence form a quadratic in tanx and solve for tanx. Remove the extraneous roots and arrive at a conclusion.