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

In any triangle ABC, tanA2tanB2tanA2+tanB2

A
aba+b
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
abc
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
aba+b+c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
ca+b
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B abc
Using formula of incentre; r=(sa)tanA2;
tanA2=rsa
Similarly; tanB2=rsb
tanA2tanB2tanA2+tanB2=rsarsbrsa+rsb
=(sb)(sa)(sb)+(sa)
=ab2sab
=abc [Since s=a+b+c2]

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