Question

If :
${ \boxed{ \bf{( \tan {}^{ - 1} x) {}^{2} + ( \cot {}^{ - 1} x) {}^{2} = \frac{5 \pi {}^{2} }{8} \: then \: find \: x.}}}$
⚠︎ Kindly Don't spamm⚠︎
– Non-copied Answer
– Well-explained Answer

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Answer 1

Given:-

${ \bf{( \tan {}^{ - 1} x) {}^{2} + ( \cot {}^{ - 1} x) {}^{2} = \frac{5 \pi {}^{2} }{8}}}$

To find:-

• The value of x.

$\huge{ \sf{ \underline{ \underline{ \color{goldenrod}{Solution:- }}}}}$

$\sf \:(tan ^{ - 1} x)^{2} + ( \frac{\pi}{2} - {tan}^{ - 1}x)^{2} = \frac{5 {\pi}^{2} }{8} \: \: \: \: \: \: \: \: \boxed{ \sf \: [tan ^{ - 1} + cot ^{ - x} = \frac{\pi}{2}]}$

$\bf \: P = {tan}^{ - 1} x$

$\boxed{ \red {\bf \: ( {P})^{2} + ( \frac{\pi}{2} - {P})^{2} = \frac{ {5\pi}^{2} }{8} }}$

$\bf \: {P}^{2} + ( \frac{\pi ^{2} }{4} + {P}^{2} - 2\pi \: l) = \frac{ {5\pi}^{2} }{8} \: \: \: \: \: \: \: \: {\boxed{(a + b) ^{2} = {a}^{2} + {b}^{2} + 2ab}}$

$\bf \: 2P ^{2} + \frac{ {\pi}^{2} }{4} - \frac{5 {\pi}^{2} }{8} - 2\pi \: l = 0$

$\bf \: 2P ^{2} - \frac{ {5\pi}^{2} }{4} + \frac{x{\pi}^{2} }{8} - \pi \: l = 0$

$\huge \purple \star \: \small{\boxed{2 P^{2} - \frac{ {3x}^{2} }{8} - \pi \:P = 0}}$

$\huge \pink \dag \: \small \boxed{16 {P}^{2} - 3\pi - 8\pi \: P = 0}$

$\boxed{16P ^{2} - 12\pi{P}+ 4\pi{P} - 3 {\pi}^{2} = 0}$

$\boxed{ \sf{4P(4P - 3\pi) + \pi(4P - 3\pi) = 0}}$

$\red{ \boxed{ \bf{(4P + x)(4P - 3\pi)}}}$

$\large \color{goldenrod}{ \boxed{{P = \frac{ - \pi}{4} and \: P = \frac{3\pi}{4} }}}$

$\bf \: if \: P = \tan ^{ - 1} x = \frac{3\pi}{4}$

$\large{ \color{aqua}{ \boxed{ \sf{x = tan \frac{3\pi}{4} }}}}$

$\large{ \color{maroon}{ \boxed{ \sf{x = tan( \frac{\pi}{2} + \frac{\pi}{4})}}}}$

$\large{ \color{blue}{{ \boxed{x = - cot \frac{\pi}{4}}}}}$

$\huge{ \red{ \boxed{ \tt{x = - 1}}}}$

Hence the value of x is -1.

Answer 2

GIVEN :–

$\\ \implies\bf(\tan^{-1} x)^{2} + (\cot^{ - 1} x)^{2} = \dfrac{5 \pi^{2} }{8}$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

$\\ \implies\bf(\tan^{-1} x)^{2} + (\cot^{ - 1} x)^{2} = \dfrac{5 \pi^{2} }{8}$

• We know that –

$\\ \implies \large \red{ \boxed{\bf\tan^{-1} x +\cot^{ - 1} x= \dfrac{\pi}{2}}}$

• So that –

$\\ \implies\bf \left( \dfrac{\pi}{2} - \cot^{-1} x \right)^{2} + (\cot^{ - 1} x)^{2} = \dfrac{5 \pi^{2} }{8}$

• Now let's put $\bf\cot^{-1} x = t$ –

$\\ \implies\bf \left( \dfrac{\pi}{2} -t\right)^{2} + (t)^{2} = \dfrac{5 \pi^{2} }{8}$

$\\ \implies\bf \dfrac{\pi ^{2} }{4} + t^{2} - 2(t) \left(\dfrac{\pi}{2} \right)+ (t)^{2} = \dfrac{5 \pi^{2} }{8}$

$\\ \implies\bf \dfrac{\pi ^{2} }{4} + t^{2} -t\pi+t^{2} = \dfrac{5 \pi^{2} }{8}$

$\\ \implies\bf \dfrac{\pi ^{2} }{4} +2t^{2} -t\pi - \dfrac{5 \pi^{2} }{8} = 0$

$\\ \implies\bf \dfrac{2\pi ^{2} - 5 {\pi}^{2}}{8}+2t^{2} -t\pi= 0$

$\\ \implies\bf - \dfrac{ 3 {\pi}^{2}}{8}+2t^{2} -t\pi= 0$

$\\ \implies\bf 16 {t}^{2}-8t\pi-3 {\pi}^{2}= 0$

$\\ \implies\bf 16 {t}^{2}-12t\pi + 4t \pi-3 {\pi}^{2}= 0$

$\\ \implies\bf 4t(4t - 3 \pi) + \pi(4t - 3 \pi)= 0$

$\\ \implies\bf (4t + \pi)(4t - 3 \pi)= 0$

$\\ \implies\bf t = -\dfrac{\pi}{4} ,\dfrac{3\pi}{4}$

• Now replace 't' –

$\\ \implies\bf \cot^{-1} x= -\dfrac{\pi}{4} ,\dfrac{3\pi}{4}$

$\\ \implies\bf x= cot \left( -\dfrac{\pi}{4} \right),cot \left(\dfrac{3\pi}{4} \right)$

$\\ \implies\bf x= - 1, - 1$

$\\ \large\implies{\boxed{\bf x= - 1}}$

• Hence , The value of 'x' is -1 .