11bhaiyya can you check this question up?
i feel tan2(θ3/2) term has to be there
then there would be some symmetry
Sides a,b,c of a triangle ABC are in AP
\cos(\theta_1)=\frac{a}{b+c},\cos(\theta_2)=\frac{b}{a+c} and \cos(\theta_3)=\frac{c}{a+b}
Then
\tan^2\left(\frac{\theta_1}{2} \right)+\tan^2\left(\frac{\theta_3}{2} \right)=?
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16 Answers
1To start with, a+c=2b
=>cos(θ2)=1/2
=>θ2=π/3
So tan2θ2/2 = 1/3
Is that beginning going to take me NEwhere[7][7]
11
39well, my idea before immediately thinking about solution
sides are in AP na yar.
phir take 3,4,5. thats called b555 triangle of substituitions :P
i have solved a lot of sums with this triangle, so kept my name :D
62subash you are riggt :)
that is why i was wondering why no one is solving :(
11a very long method:
take sides as r-d,r,r+d
and converting
tan2θ to sec2θ-1
and using cos2(θ1/2)=1+cosθ1/2
we can also get the result
11but the second method is fool proof and valid for a subjective one even :)
62oh ok
that is what i thought.. how did prophet sir's hint throw up the equality :D
341What I was trying to say was, we have in a triangle
\cos A + \cos B + \cos C \le \frac{3}{2}
whereas Nesbitt's Inequality says that \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ge \frac{3}{2}
Equality in the first case occurs when A=B=C = 60o
and in the second case when a =b=c.
11@the prophet sir( or anyone willing to explain)
isnt the inequality used by you valid only when A,B,C are the angles of a triangle
i maybe wrong please bear with me and explain [78]
341I completely misread the question :D
To redeem myself, i will write out the solution.
\cos \theta_1 = 2 \cos^2 \frac {\theta_1}{2} - 1 = \frac{a}{b+c} \Rightarrow \sec^2 \frac {\theta_1}{2} = \frac{2(b+c)}{a+b+c}
Similarly, \sec^2 \frac {\theta_2}{2} = \frac{2(a+b)}{a+b+c}
Hence, \tan^2 \frac {\theta_1}{2}+ \tan^2 \frac {\theta_2}{2} = \sec^2 \frac {\theta_2}{2}+ \sec^2 \frac {\theta_2}{2} -2 = \frac{2(a+2b+c)}{a+b+c} - 2
Since a+c = 2b, this simplifies to 2