If `(a^(2)+1)/(2a)=costheta, then (a^(6)+1)/(2a^(3))=`

If x+1/x=2costheta then x^3+1/x^3=

If sqrtx+(1)/(sqrt(x))=2costheta then x^(6)+x^(-6)=

If x+1/x=2costheta , then x^3+1/x^3=

If sqrtx+1/sqrtx=2costheta , then x^6+x^(-6)=

sintheta+costheta=(sqrt(3)+1)/(2)

Sides a , b , c of !ABC are in A.P. and costheta_(1)=a/(b+c)costheta_(2)=b/(a+c),costheta_(3)=c/(a+b) , then tan^(2)(theta_(1))/2+tan^(2)(theta_(3))/2 =

Sides a , b , c of !ABC are in A.P. and costheta_(1)=a/(b+c)costheta_(2)=b/(a+c),costheta_(3)=c/(a+b) , then tan^(2)(theta_(1))/2+tan^(2)(theta_(3))/2 =

(2+sqrt(3))costheta =1 - sintheta

Prove the following: ((tantheta+(1)/(costheta))^2+((tantheta-(1)/(costheta))^2=2(frac(1+sin^2theta)(1-sin^2theta))

If costheta_1 =2costheta_2, then tan((theta_1-theta_2)/2)tan((theta_1+theta_2)/2) is equal to (a) 1/3 (b) -1/3 (c) 1 (d) -1