wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Inverse circular functions,Principal values of sin1x,cos1x,tan1x.
tan1x+tan1y=tan1x+y1xy, xy<1
π+tan1x+y1xy, xy>1.
(a) If sin1x+sin1y+sin1z=3π/2, then the value of
x100+y100+z1009x101+y101+z101 is

(b) sin1axc+sin1bxc=sin1x
where a2+b2=c2,c0.

Open in App
Solution

(a) . We know that sin1xπ/2
Hence from the given relation we observe that each of sin1x,sin1y and sin1z will be π/2 so that x=y=z=sin(π/2)=1 393=0
(b) sin1axc1b2x2c2+bxc1a2x2c2=sin1x
x[ac2b22x+bc2a2x2c2]=0
x=0 is one solution which satisfies the given equation.
Again ac2b2x2+bc2a2x2=c2
Squaring both sides,
a2(c2b2x2)+b2(c2a2x2)+2ab=c4
c2(a2+b2)2a2b2x2+2ab=c4
Put a2+b2=c2 and cancel c4 from both sides
abx2=c2b2x2c2a2x2
Square again
a2b2x4=c4c2(a2+b2)x2+a2b2x4
or 0=c4(1x2)x=±1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Property 5
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon