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Find the value of ${\cos ^{ - 1}}\left( {\dfrac{{{a^{ - x}} - {a^x}}}{{{a^{ - x}} + {a^x}}}} \right)$
Answer
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Verified
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Hint: - Use the identity ${a^x} = \tan \theta $
Given equation is
$
{\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 ${a^x} = \tan \theta ............\left( 1 \right)$
$ \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 $\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos 2\theta $
$ \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 ${\cos ^{ - 1}}x$ will always lie between $\left( {0,\pi } \right)$
$
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
$
{a^x} = \tan \theta \\
\Rightarrow \theta = {\tan ^{ - 1}}\left( {{a^x}} \right) \\
$
From equation 2
$
\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 $\tan 0 = 0$ and $\tan \dfrac{\pi }{2} = \infty $
Therefore from above equation
$
\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 ${a^x} = \tan \theta $, then simplify using some basic trigonometric properties which is stated above, then always remember the domain of ${\cos ^{ - 1}}x$, then again simplify we will get the required answer.
Given equation is
$
{\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 ${a^x} = \tan \theta ............\left( 1 \right)$
$ \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 $\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos 2\theta $
$ \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 ${\cos ^{ - 1}}x$ will always lie between $\left( {0,\pi } \right)$
$
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
$
{a^x} = \tan \theta \\
\Rightarrow \theta = {\tan ^{ - 1}}\left( {{a^x}} \right) \\
$
From equation 2
$
\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 $\tan 0 = 0$ and $\tan \dfrac{\pi }{2} = \infty $
Therefore from above equation
$
\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 ${a^x} = \tan \theta $, then simplify using some basic trigonometric properties which is stated above, then always remember the domain of ${\cos ^{ - 1}}x$, then again simplify we will get the required answer.
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