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  • <img alt="\cos ^{-1}(\cos (-5))+\sin ^{-1}(\sin (6))-\tan ^{-1}(\tan (12))" src="https://entrancecorner.codecogs.com/gif.latex?%5Ccos%20%5E%7B-1%7D%28%5Ccos%20%28-5%29%29&plus;%5Csin%20%5E%7B-1%7D%28%5Csin%20%286%29%29-%5Ctan%20%5E%7B-1%7D%28%5Ctan%20

\cos ^{-1}(\cos (-5))+\sin ^{-1}(\sin (6))-\tan ^{-1}(\tan (12)) is equal to : (The inverse trigonometric functions take the principal values)
 
Option: 1 3 \pi+1
Option: 2 3 \pi-11
Option: 3 4 \pi-11
Option: 4 4 \pi-9

Answers (1)

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\cos^{-1}\left ( \cos \left ( -5 \right ) \right )+\sin^{-1}\left ( \sin \left ( 6 \right ) \right )-\tan^{-1}\left ( \tan \left ( 12 \right ) \right )= \cos^{-1}\left ( \cos \left ( 5 \right ) \right )+\sin^{-1}\left ( \sin \left ( 6 \right ) \right )-\tan^{-1}\left ( \tan \left ( 12 \right ) \right )

Using graphs of y= T^{-1}\left ( T\left ( x \right ) \right )

= \left ( 2\pi -5 \right )+\left ( 6-2\pi \right )-\left ( 12-4\pi \right )
= 4\pi -11

Posted by

Kuldeep Maurya

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