Statement 1. If `p=7+tanalpha*tanbeta, q=5+tanbeta*tangamma` and `r=3=tangamma*tanalpha` then the maximum value of `sqrt(p)+sqrt(q)+sqrt(r)` is 4., Statement 2. `tanalpha.tanbeta+tanbeta.tangamma+tangamma*tanalpha=1`

If tanalpha+tanbeta+tangamma=tanalphatanbetatangamma,then

If tanalpha+tanbeta+tangamma=tanalphatanbetatangamma,then

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

Let alpha,betaa n dgamma be some angles in the first quadrant satisfying tan(alpha+beta)=(15)/8a n dcos e cgamma=(17)/8, then which of the following hold(s) good? (a) alpha+beta+gamma=pi (b)cotalpha+tanbeta+tangamma=tanalphatanbetatangamma (c)tanalpha+tanbeta+tangamma=tanalphatanbetatangamma (d)tanalphatanbeta+tanbetatangamma+tangammatanalpha=1

Let alpha,betaa n dgamma be some angles in the first quadrant satisfying tan(alpha+beta)=(15)/8a n dcos e cgamma=(17)/8, then which of the following hold(s) good? alpha+beta+gamma=pi cotalpha+tanbeta+tangamma=tanalphatanbetatangamma tanalpha+tanbeta+tangamma=tanalphatanbetatangamma tanalphatanbeta+tanbetatangamma+tangammatanalpha=1

If alpha+beta+gamma=2pi, then- a.tanalpha/2+(tan beta)/2+tangamma/2=tanalpha/2tanbeta/2tangamma/2 b. tanalpha/2+tanbeta/2+tan"(beta/2)tangamma/2=tangamma/2tanalpha/2=1 c.tanalpha/2+t a nbeta/2+tangamma/2=tanalpha/2tanbeta/2tangamma/2 d.tan(alpha/2)tan(beta/2) +tan(gamma/2)tan(beta/2) +tan(gamma/2)tan(alpha/2) =0

(1+tanalphatanbeta)^2+(tanalpha-tanbeta)^2=

(1+tanalphatanbeta)^2+(tanalpha-tanbeta)^2=

(1+tanalphatanbeta)^2+(tanalpha-tanbeta)^2=