wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

If x1,x2,x3 are positive roots of x36x2+3px2p=0(pR), then the value of sin1(1x1+1x2)+cos1(1x2+1x3)tan1(1x3+1x1) is equal to

A
π8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
π6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
π4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
π
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C π4
x36x2+3px2p=0
x1
x2
x3
are the zeros
x1+x2+x3=6
x1x2+x2x3+x3x1=3p x1+x2+x33(2p)13,2(2p)13p4
x1x2x3=2p
3p3((x1x2x3)2)1/3
p34p2
p4
p=4
x1x2+x2x3+x3x1=12
x1x2x3=8
x1+x2+x3=6
sin1(1)+cos1(1)tan1(1)
π2+0π4=π4.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Basic Inverse Trigonometric Functions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon