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Question: Find the value of \({\cos ^{ - 1}}\left( {\dfrac{{{a^{ - x}} - {a^x}}}{{{a^{ - x}} + {a^x}}}} \right...

Find the value of cos⁡−1(a−x−axa−x+ax){\cos ^{ - 1}}\left( {\dfrac{{{a^{ - x}} - {a^x}}}{{{a^{ - x}} + {a^x}}}} \right)

Explanation

Solution

Hint: - Use the identity ax=tan⁡θ{a^x} = \tan \theta
Given equation is
cos⁡−1(a−x−axa−x+ax) =cos⁡−1(1ax−ax1ax+ax)=cos⁡−1(1−a2x1+a2x)  {\cos ^{ - 1}}\left( {\dfrac{{{a^{ - x}} - {a^x}}}{{{a^{ - x}} + {a^x}}}} \right) \\\ = {\cos ^{ - 1}}\left( {\dfrac{{\dfrac{1}{{{a^x}}} - {a^x}}}{{\dfrac{1}{{{a^x}}} + {a^x}}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{1 - {a^{2x}}}}{{1 + {a^{2x}}}}} \right) \\\
Let ax=tan⁡θ............(1){a^x} = \tan \theta ............\left( 1 \right)
⇒cos⁡−1(1−a2x1+a2x)=cos⁡−1(1−(tan⁡θ)21+(tanθ)2)=cos⁡−1(1−tan⁡2θ1+tan⁡2θ)\Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{1 - {a^{2x}}}}{{1 + {a^{2x}}}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{1 - {{\left( {\tan \theta } \right)}^2}}}{{1 + {{\left( {tan\theta } \right)}^2}}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)
Now, as we know 1−tan⁡2θ1+tan⁡2θ=cos⁡2θ\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos 2\theta
⇒cos⁡−1(1−tan⁡2θ1+tan⁡2θ)=cos⁡−1(cos⁡2θ)=2θ\Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) = {\cos ^{ - 1}}\left( {\cos 2\theta } \right) = 2\theta
But as we know cos⁡−1x{\cos ^{ - 1}}x will always lie between (0,π)\left( {0,\pi } \right)
0⩽cos⁡−1(cos⁡2θ)⩽π ⇒0⩽2θ⩽π ⇒0⩽θ⩽π2...........(2)  0 \leqslant {\cos ^{ - 1}}\left( {\cos 2\theta } \right) \leqslant \pi \\\ \Rightarrow 0 \leqslant 2\theta \leqslant \pi \\\ \Rightarrow 0 \leqslant \theta \leqslant \dfrac{\pi }{2}...........\left( 2 \right) \\\
Now from equation 1
ax=tan⁡θ ⇒θ=tan⁡−1(ax)  {a^x} = \tan \theta \\\ \Rightarrow \theta = {\tan ^{ - 1}}\left( {{a^x}} \right) \\\
From equation 2
⇒0⩽tan⁡−1(ax)⩽π2 ⇒tan⁡0⩽ax⩽tan⁡π2  \Rightarrow 0 \leqslant {\tan ^{ - 1}}\left( {{a^x}} \right) \leqslant \dfrac{\pi }{2} \\\ \Rightarrow \tan 0 \leqslant {a^x} \leqslant \tan \dfrac{\pi }{2} \\\
As we know the value of tan⁡0=0\tan 0 = 0 and tan⁡π2=∞\tan \dfrac{\pi }{2} = \infty
Therefore from above equation
tan⁡0⩽ax⩽tan⁡π2 =0⩽ax⩽∞  \tan 0 \leqslant {a^x} \leqslant \tan \dfrac{\pi }{2} \\\ = 0 \leqslant {a^x} \leqslant \infty \\\
So, this is the required solution.

Note: -In such types of questions first substitute ax=tan⁡θ{a^x} = \tan \theta , then simplify using some basic trigonometric properties which is stated above, then always remember the domain of cos⁡−1x{\cos ^{ - 1}}x, then again simplify we will get the required answer.