Download the app

Questions  

Sec11+x2+Cosece11+y2y+Cot11z=π then

a
x+y+z=0
b
x+y+z=1
c
x+y+z=xyz
d
x+y+z=− xyz

detailed solution

Correct option is C

Let x=tanA, y=tanB and z=tanC.   Then sec-11+x2+cosec-11+y2y+cot-11z=π ⇒sec-11+tan2A+cosec-11+tan2BtanB+cot-11tanC=π ⇒sec-1secA+cosec-1cosecB+cot-1cotC=π ⇒A+B+C=π ⇒tanA+tanB+tanC=tanAtanBtanC ⇒x+y+z=xyz

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

The value of x for which SinCot11+x=CosTan1x is


phone icon
whats app icon