Question

Find the value of tan^-1(1)+cos^-1(-1/2)+sin^-1(-1/2) Please answer it fast,correct answer will be brainliest

Answer 1

Answer:

$\dfrac{3\pi}{4}$

Explanation:

$\sf{Let\ tan}^{-2}(1)=x$

$\sf{Then\ tan\ x=0=1=tan\dfrac{\pi}{4}$

$\because \sf{tan}^{-2}(1)=\dfrac{\pi}{4}$

$\sf{Let\ cos^{-1}\bigg\lgroup-\dfrac{1}{2}\bigg\rgroup=y$

$\sf{Then, cos\ y}=-\dfrac{1}{2}=\sf{-cos}\bigg\lgroup{\dfrac{\pi}{3}\bigg\rgroup}=\sf{cos}{-\pi \dfrac{\pi}{3}=cos\bigg\lgroup\dfrac{2x}{3}\bigg\rgroup$

$\because\ \sf{cos^{-2}\bigg\lgroup-\dfrac{1}{2}\bigg\rgroup=\dfrac{2\pi}{3}$

$\sf{Let\ sin^{-1}\bigg\lgroup-\dfrac{1}{2}\bigg\rgroup=z$

$\sf{Then,sin\ z=-\dfrac{1}{2}=-sin\bigg\lgroup\dfrac{\pi}{6}\bigg\rgroup=-sin\bigg\lgroup-\dfrac{\pi}{6}\bigg\rgroup$

$\because\ \sf{sin}^{-1}\bigg\lgroup-\dfrac{1}{2}\bigg\rgroup=\dfrac{-\pi}{6}$

$\because\ \sf{tan}^{-1}(1)+\sf{cos}^{-1}\bigg\lgroup-\dfrac{1}{2}\bigg\rgroup+\sf{sin}^{-1}\bigg\lgroup-\dfrac{1}{2}\bigg\rgroup$

$\rightarrow \dfrac{\pi}{4}+\dfrac{2\pi}{3}=\dfrac{\pi}{6}$

$\rightarrow \dfrac{3\pi+8\pi-2\pi}{12}$

$\rightarrow \dfrac{9\pi}{12}$

$\rightarrow \dfrac{3\pi}{4}$

Answer 2

Solution :

$\sf \mapsto { \tan}^{ - 1} (1) + { \cos}^{ - 1} ( - \frac{1}{2} ) + { \sin}^{ - 1} ( - \frac{1}{2} ) \\ \\ \sf \mapsto \frac{\pi}{4} +( \pi - \frac{\pi}{3} ) + ( - \frac{\pi}{6} ) \\ \\ \sf \mapsto \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{3} \\ \\ \sf Taking \: LCM \: , \: we \: get \\ \\ \sf \mapsto \frac{2\pi + 8\pi - 2\pi}{12} \\ \\ \sf \mapsto \frac{9\pi}{12} \\ \\ \sf \mapsto \frac{3\pi}{4}$

Remmember :

$\sf \star \: \: { \sin}^{ - 1} ( - x) = - { \sin}^{ - 1} ( x) \\ \\\sf \star \: \: { \cos}^{ - 1} ( - x) = \pi - { \cos}^{ - 1} ( x) \\ \\ \sf \star \: \: { \tan}^{ - 1} ( - x) = - { \tan}^{ - 1} ( x) \\ \\ \sf \star \: \: { \cot}^{ - 1} ( - x) = \pi - { \cot}^{ - 1} ( x) \\ \\\sf \star \: \: { \sec}^{ - 1} ( - x) = \pi - { \sec}^{ - 1} ( x) \\ \\ \sf \star \: \: {cosec}^{ - 1} ( - x) = - {cosec}^{ - 1} ( x)$