# Find the value of ${\cos ^{ - 1}}\left( {\dfrac{{{a^{ - x}} - {a^x}}}{{{a^{ - x}} + {a^x}}}} \right)$

Answer

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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.

Last updated date: 24th Sep 2023

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