Question

Q19. If (tan⁻¹(x))² + (cot⁻¹(x))² = 5π²/8, then find x.​

Answer 1

Topic :-

Inverse Trigonometric Function

Given :-

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

To Find :-

Value of 'x'.

Solution :-

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

$\sf{(\tan^{-1}x)^2+\left(\dfrac{\pi}{2}-\tan^{-1}x\right)^2=\dfrac{5\pi^2}{8}}$

$\sf{\left(\because \tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2},when\:x\in R\right)}$

$\sf{Put\:\tan^{-1}x=p,}$

$\sf{p^2+\left(\dfrac{\pi}{2}-p\right)^2=\dfrac{5\pi^2}{8}}$

$\sf{p^2+\left(\dfrac{\pi}{2}\right)^2-2\cdot\left( \dfrac{\pi}{2}\right)\cdot p+p^2=\dfrac{5\pi^2}{8}}$

$\sf{(\because (a-b)^2=a^2-2ab+b^2)}$

$\sf{p^2+p^2-\not{2}\cdot\left( \dfrac{\pi}{\not{2}}\right)\cdot p+\left(\dfrac{\pi}{2}\right)^2-\dfrac{5\pi^2}{8}=0}$

$\sf{2p^2- p\pi+\dfrac{\pi^2}{4}-\dfrac{5\pi^2}{8}=0}$

$\sf{\dfrac{8(2p^2)- 8p\pi+2\pi^2-5\pi^2}{8}=0}$

$\sf{\dfrac{16p^2- 8p\pi-3\pi^2}{8}=0}$

$\sf{16p^2- 8p\pi-3\pi^2=0}$

Applying Quadratic Formula,

$\sf{p=\dfrac{-(-8\pi)\pm \sqrt{(-8\pi)^2-4(16)(-3\pi^2)}}{32}}$

$\sf{p=\dfrac{8\pi\pm \sqrt{64\pi^2+192\pi^2}}{32}}$

$\sf{p=\dfrac{8\pi\pm \sqrt{256\pi^2}}{32}}$

$\sf{p=\dfrac{8\pi\pm 16\pi}{32}}$

$\sf{p=\dfrac{8\pi+16\pi}{32}\:\:or\:\:\dfrac{8\pi-16\pi}{32}}$

$\sf{p=\dfrac{24\pi}{32}\:\:or\:\:\dfrac{-8\pi}{32}}$

$\sf{ p = \dfrac{ 3\pi}{4} \: \: or \: \: \dfrac{ - \pi}{4}}$

Put back value of p,

$\sf{\tan^{-1}x=\dfrac{3\pi}{4}\:\:or\:\:\dfrac{-\pi}{4}}$

$\sf{x=\tan \left(\dfrac{3\pi}{4}\right)\:\:or\:\:\tan\left(\dfrac{-\pi}{4}\right)}$

$\sf{x=-1}$

$\sf{\left(\because\tan\left( \dfrac{3\pi}{4}\right)=\tan\left( \dfrac{-\pi}{4}\right)=-1\right)}$

Answer :-

$\sf{Value\:of\:\:\bold{x=-1}}$

Answer 2

Step-by-step explanation:

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