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

In any ABC, tanA2-tanB2tanA2+tanB2 is equal to:


A

a-ba+b

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

a-bc

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

ca-b

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

None of these.

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

a-bc


Explanation for the correct option.

Step 1: Simplify tanA2-tanB2tanA2+tanB2.

tanA2-tanB2tanA2+tanB2=sinA/2cosA/2-sinB/2cosB/2sinA/2cosA/2+sinB/2cosB/2=sinA2·cosB2-sinB2·cosB2sinA2·cosB2+sinB2·cosB2=sinA-B2sinA+B2[sin(x-y)=sinxcosy-sinycosxandsinx+y=sinxcosy+sinycosxand]

By angle sum property of a triangle,

A+B+C=180°A+B2=90°-C2

So,

sinA-B2sin90°-C2=sinA-B2cosC2[Bysin(90°-θ)=cosθ]...(1)

Multiplying the numerator and denominator of 1 by 2sinC2, we get

2sinC2sinA-B22sinC2cosC2=2sin90°-A+B2sinA-B2sin2C2[sin2θ=2sinθcosθ]=2cosA+B2sinA-B2sinC[sin90°-θ=cosθ]=sinA-sinBsinC...2[sinx-siny=2cosx+y2sinx-y2]

We know that asinA=bsinB=csinC=k, so 2 will be

=ak-bkck=a-bc

Hence, option B is correct.


flag
Suggest Corrections
thumbs-up
8
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
General Solutions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon