Q1 If roots of x3-px2-r=0 are tanα,tanβ,tanγ ,find value of sec2α.sec2β.sec2γ
I did by converting secx into tanx and then using equation to solve ......any better method ??
Q2 Prove that sinxcos3x+sin3xcos9x+sin9xcos27x=12[tan27x-tanx]
1L.H.S = \frac{2Sinx*Cosx}{2Cos3x*Cosx}+ \frac{2Sin3x*cos3x}{2cos9x*cos3x}+\frac{2sin9x*cos9x}{2cos27x*cos9x}
L.H.S = \frac{sin(3x-x)}{2sin3x*Cosx}+ \frac{sin(9x-3x)}{2cos9x*cos3x}+\frac{sin(27x-9x)}{2cos27x*cos9x}
L.H.S = 1/2(tan3x-tanx+tan9x-tan3x+tan27x-tan9x)= 1/2(tan27x-tanx)
using sin2a=2sina*cosa and sin(a-b) = sina*cosb-sinb*cosa
341If m is a root of x^3-px^2-r = 0 ,then we have
m^3-pm^2-r = 0 \Rightarrow m^3 = pm^2+r \Rightarrow m^6 - (pm^2+r)^2 = 0
Setting t = m2, t is a root of the equation z^3-(pz+r)^2 = 0
From this we see that \tan^2 \alpha, \tan^2 \beta, \tan^2 \gamma are the roots of the above cubic.
Now
(1+\tan^2 \alpha) (1+ \tan^2 \beta)(1+ \tan^2 \gamma) \\ \\ = -(-1-\tan^2 \alpha) (-1- \tan^2 \beta)(-1- \tan^2 \gamma)= -P(-1)
where P(z) = z^3-(pz+r)^2
Then we have -P(-1) = (1+(p-r)^2)
