If `alpha+beta+gamma=2pi`, then show that `tan.alpha/2 + tan.beta/2 + tan.gamma/2 = tan.alpha/2 tan.beta/2 tan.gamma/2`.

If 2tanalpha=3 tanbeta , show that tan(alpha-beta)=(sin2beta)/(5-cos2beta)

If alpha+beta=pi/2a n dbeta+gamma=alpha, then tanalpha equals

If 2 sin alphacos beta sin gamma=sinbeta sin(alpha+gamma),then tan alpha,tan beta and tan gamma are in

Let alpha,beta,gamma > 0 and alpha+beta+gamma=pi/2. Statement-1: |tan alpha tan beta-(a!)/6|+|tan beta tan gamma-(b!)/2|+|tan gamma tanalpha-(c!)/3| le 0, where n! =1.2..........n, then tan alpha tanbeta,tanbeta tangamma, tan gamma tan alpha=1 Settlement 2 : tan alpha tanbeta+,tanbeta tangamma+, tan gamma tan alpha=1

lf cos^2 alpha -sin^2 alpha = tan^2 beta , then show that tan^2 alpha = cos^2 beta-sin^2 beta .

If tan (alpha- beta)= (sin 2 beta)/(5-cos 2 beta) , find tan alpha: tan beta .

If alpha and beta satisfying 2 sec2 alpha = tan beta + cot beta , then alpha+beta =

If [[alpha, beta], [gamma, -alpha]] is to be square root of two-rowed unit matrix, then alpha,beta and gamma should satisfy the relation. a. 1-alpha^2+betagamma=0 b. alpha^2+betagamma=0 c. 1+alpha^2+betagamma=0 d. 1-alpha^2-betagamma=0