# Trigonometry terms in GP and their values

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

$$\dfrac{\sin\pi/12\cos(\pi/12-x)}{\cos\pi/12\sin(\pi/12-x)}=?$$

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

• ?? I couldn't reach to this expression. Wasn't it simple manipulations? – CodeBlooded Nov 27 '17 at 14:21
• @Khizir, If $a,b,c$ are in GP $$b/a=c/b$$ Now as already mentioned please write tan as $\sin/\cos$ – lab bhattacharjee Nov 27 '17 at 14:26
• Yes, I know this thing, but taking it as sin/cos doesn't bring up this expression for me. – CodeBlooded Nov 27 '17 at 14:27
• @Khizir, Please share your progress – lab bhattacharjee Nov 27 '17 at 14:29
• 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.. :) – CodeBlooded Nov 27 '17 at 14:32

Hint:

Writing like $b^2=ca,$

$$\dfrac{\sin^2A}{\cos^2A}=\dfrac{\sin(A-x)\sin(A+x)}{\cos(A-x)\cos(A+x)}$$

and

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

Finally double formula for cosine.

Please let me know if you face any further problem?