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Question: \({{\cos }^{-1}}\left( \cos \left( 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right) \right)\) is equ...

cos⁡−1(cos⁡(2cot⁡−1(2−1))){{\cos }^{-1}}\left( \cos \left( 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right) \right) is equal to:
[a] 2−1\sqrt{2}-1
[b] π4\dfrac{\pi }{4}
[c] 3π4\dfrac{3\pi }{4}
[d] None of the above.

Explanation

Solution

Use the fact that tan⁡(π8)=2−1\tan \left( \dfrac{\pi }{8} \right)=\sqrt{2}-1 and then use the identities cot⁡x=tan⁡(π2−x),cot⁡−1(cot⁡x)=x+nπ\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+nπx+n\pi is within the interval (0,π)\left( 0,\pi \right). Alternatively use cos⁡−1(cos⁡x)=x+2mπ{{\cos }^{-1}}\left( \cos x \right)=x+2m\pi , where m is suitably chosen so that the value of x+2mπx+2m\pi is within the interval [0,π]\left[ 0,\pi \right] and the assume the expression be equal to y. Take cot on both sides and then use cot⁡2x=cot⁡2x−12cot⁡x\cot 2x=\dfrac{{{\cot }^{2}}x-1}{2\cot x}. Simplify to get the expression for coty. Find in the interval (0,π)\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⁡(π8)=2−1\tan \left( \dfrac{\pi }{8} \right)=\sqrt{2}-1
Hence, we have tan⁡(π2−3π8)=2−1\tan \left( \dfrac{\pi }{2}-\dfrac{3\pi }{8} \right)=\sqrt{2}-1
We know that cot⁡x=tan⁡(π2−x)\cot x=\tan \left( \dfrac{\pi }{2}-x \right)
Put x=3π8x=\dfrac{3\pi }{8}, we have
cot⁡(3π8)=tan⁡(π2−3π8)\cot \left( \dfrac{3\pi }{8} \right)=\tan \left( \dfrac{\pi }{2}-\dfrac{3\pi }{8} \right)
Hence, we have
cot⁡(3π8)=2−1\cot \left( \dfrac{3\pi }{8} \right)=\sqrt{2}-1
Hence, we have
cot⁡−1(2−1)=cot⁡−1(cot⁡(3π8))=3π8+nπ{{\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 3π8∈(0,π)\dfrac{3\pi }{8}\in \left( 0,\pi \right), we have n = 0.
Hence cot⁡−1(2−1)=3π8{{\cot }^{-1}}\left( \sqrt{2}-1 \right)=\dfrac{3\pi }{8}
Hence, we have cos⁡(2cot⁡−1(2−1))=cos⁡(23π8)=cos⁡(3π4)=cos⁡(π−π4)\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⁡(π4)=12\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}} and cos⁡(π−x)=−cos⁡x\cos \left( \pi -x \right)=-\cos x
Put x=π4x=\dfrac{\pi }{4}, we get
cos⁡(π−π4)=−cos⁡(π4)=−12\cos \left( \pi -\dfrac{\pi }{4} \right)=-\cos \left( \dfrac{\pi }{4} \right)=\dfrac{-1}{\sqrt{2}}
Hence , we have cos⁡(2cot⁡−1(2−1))=−12\cos \left( 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right)=-\dfrac{1}{\sqrt{2}}
Now, we know that arccos⁡(−x)=π−arccos⁡x\arccos \left( -x \right)=\pi -\arccos x
Hence, we havecos⁡−1(cos⁡(2cot⁡−1(2−1)))=cos⁡−1(−12)=π−cos⁡−1(12){{\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⁡(π4)=12\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}} , we have arccos⁡(12)=π4\arccos \left( \dfrac{1}{\sqrt{2}} \right)=\dfrac{\pi }{4}
Hence, we have
cos⁡−1(cos⁡(2cot⁡−1(2−1)))=π−π4=3π4{{\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(cos⁡x)=x+2mπ{{\cos }^{-1}}\left( \cos x \right)=x+2m\pi , where m is suitably chosen so that the value of x+2mπx+2m\pi is within the interval [0,π]\left[ 0,\pi \right].
Hence, we get
cos⁡−1(cos⁡(2cot⁡−1(2−1)))=2mπ+2cot⁡−1(2−1){{\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=2mπ+2cot⁡−1(2−1)y=2m\pi +2{{\cot }^{-1}}\left( \sqrt{2}-1 \right)
Taking cot on both sides, we get
cot⁡y=cot⁡(2mπ+2cot⁡−1(2−1))\cot y=\cot \left( 2m\pi +2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right)
Now since cotx is periodic with period π\pi , we have
cot⁡y=cot⁡(2cot⁡−1(2−1))\cot y=\cot \left( 2{{\cot }^{-1}}\left( \sqrt{2}-1 \right) \right)
Using cot⁡2x=cot⁡2x−12cot⁡x\cot 2x=\dfrac{{{\cot }^{2}}x-1}{2\cot x} , we get
cot⁡y=cot⁡2(cot⁡−1(2−1))−12cot⁡(cot⁡−1(2−1))\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⁡(cot⁡−1x)=x\cot \left( {{\cot }^{-1}}x \right)=x
Hence, we have
cot⁡y=(2−1)2−12(2−1)=2−222(2−1)=−1\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=3π4y=\dfrac{3\pi }{4}, which is the same as obtained above.
[2] You can prove that tan⁡(π8)=2−1\tan \left( \dfrac{\pi }{8} \right)=\sqrt{2}-1, by putting x=π4x=\dfrac{\pi }{4} in the identity tan⁡2x=2tan⁡x1−tan⁡2x\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.