Question: If tan((π/12) - x), tan (π/12), tan((π/12) + x) in the order are the three consecutive terms of a GP then sum all the solutions in [0,314] is kπ. Find value of k.

Attempt: I tried assuming a = tan (π/12) and y = tanx to make my calculations easier. a^2 = (a+y)/(1-ay) * (a-y)/(1+ay)

Simplifying this simple removes y from the expression. What should I do? Where am I missing.


Write tan as $\dfrac{\sin}{\cos}$ like


Now apply componendo & dividendo(https://brilliant.org/wiki/componendo-and-dividendo/)

  • $\begingroup$ ?? I couldn't reach to this expression. Wasn't it simple manipulations? $\endgroup$ – CodeBlooded Nov 27 '17 at 14:21
  • $\begingroup$ @Khizir, If $a,b,c$ are in GP $$b/a=c/b $$ Now as already mentioned please write tan as $\sin/\cos$ $\endgroup$ – lab bhattacharjee Nov 27 '17 at 14:26
  • $\begingroup$ Yes, I know this thing, but taking it as sin/cos doesn't bring up this expression for me. $\endgroup$ – CodeBlooded Nov 27 '17 at 14:27
  • $\begingroup$ @Khizir, Please share your progress $\endgroup$ – lab bhattacharjee Nov 27 '17 at 14:29
  • $\begingroup$ Okay okay I get that... I was using b^2 = a*c. I thought finally we will have to solve it, why not directly use it... Thanks.. :) $\endgroup$ – CodeBlooded Nov 27 '17 at 14:32


Writing like $b^2=ca,$


Now use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $


Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

Then https://brilliant.org/wiki/componendo-and-dividendo/

Finally double formula for cosine.

Please let me know if you face any further problem?


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