Prove that: `tan(alpha-beta)+tan(beta-gamma)+tan (gamma-alpha) = tan(alpha-beta) tan (beta-gamma) tan (gamma-alpha)`.

tan(alpha-beta)+tan(beta-alpha)+tan(gamma-alpha)=tan(alpha-beta)tan(beta-gamma)tan(gamma-alpha) Reason (R):In Delta ABC sum tan A=pi tan A

If alpha + beta + gamma = pi , show that : tan (beta+gamma-alpha) + tan (gamma+alpha-beta) + tan (alpha+beta-gamma) = tan (beta+gamma-alpha) tan (gamma+alpha-beta) tan (alpha+beta-gamma)

If alpha + beta + gamma = pi , show that : tan (beta+gamma-alpha) + tan (gamma+alpha-beta) + tan (alpha+beta-gamma) = tan (beta+gamma-alpha) tan (gamma+alpha-beta) tan (alpha+beta-gamma)

If alpha+beta+gamma=2pi, then (a) tan(alpha/2)+tan(beta/2)+tan(gamma/2)=tan(alpha/2)tan(beta/2)tan(gamma/2) (b)tan(alpha/2)tan(beta/2)+tan(beta/2)tan(gamma/2)+tan(gamma/2)tan(alpha/2)=1 (c)tan(alpha/2)+tan(beta/2)+tan(gamma/2)=-tan(alpha/2)tan(beta/2)tan(gamma/2) (d)none of these

If (sec alpha+tan alpha)(sec beta+tan beta)(sec gamma+tan gamma)=tan alpha tan beta tan gamma, then (sec alpha-tan alpha)(sec beta-tan beta)(sec gamma-tan gamma)=

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

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

If alpha+beta+gamma=2pi, then (a) tan(alpha/2)+tan(beta/2)+tan(gamma/2)=tan(alpha/2)tan(beta/2)tan(gamma/2) (b) tan(alpha/2)tan(beta/2)+tan(beta/2)tan(gamma/2)+tan(gamma/2)tan(alpha/2)=1 (c) tan(alpha/2)+tan(beta/2)+tan(gamma/2)=-tan(alpha/2)tan(beta/2)tan(gamma/2) (d)none of these