Expert: Vijilant - 10/31/2009

Question
Sir, please solve these problems soon,

If cos2B=cos(A+C)/cos(A-C) then show that tanA,tanB and tanC are in G.P.

Prove that (1/cos0°+cos1°)+(1/cos1° +cos2°)+.......+ (1/cos88° +cos90°)= cos1°/sin^2 1°

Find all x between -pi/2 and pi/2 such that 1-sin^4x-cos^2x=1/16

Evaluate sin{n.pi+(-1)^n.(pi/4)}, where n is an integer

If 3tanΘtanØ1 then prove that 2cos(Θ+Ø)=cos(Θ-Ø)

Thanks a lot!

Answer
Hello Sneha
This is not my expertise, but I will help you with some of them.
cos(2B) = 2 cos^2(B) - 1 = 2/sec^2(B) -1 = (1-tan^2(B))/(1+tan^2(B))
Then, clearing the fraction:
cos(A-C) -cos(A-C)tan^2(B) = cos(A+C) +cos(A+C)tan^2(B)
tan^2(B)(cos(A+C)+cos(A-C)) = cos(A-C)-cos(A+C)
tan^2(B)*2cos(A)cos(C) = 2sin(A)sin(C)
tan^2(B) = tan(A)tan(C) provingthey are in GP.

In the third one, use 1-cos^2(x) = sin^2(x) and you have a quadratic in sin^2(x).
Solve this to give sin^2(x) = (2+/- 3^(1/2))/4.
The rest is easy.

The fourth one.  Consider odd and even n separately and show that in both cases you get 2^(-1/2).

The other 2 contain misprints.

Best wishes

vijilant