0
$\begingroup$

Question:

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.

$\endgroup$
0
$\begingroup$

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/)

$\endgroup$
  • $\begingroup$ ?? 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
0
$\begingroup$

Hint:

Writing like $b^2=ca,$

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

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

and

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.