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]
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
1
11bhaiyya can you check this question up?
i feel tan2(θ3/2) term has to be there
then there would be some symmetry
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