`(xtanalpha+ycotalpha)(xcotalpha+ytanalpha)-4xycos^2 2alpha=`

If alpha = (2pi )/(7) , then tan alpha tan 2 alpha + tan 2 alpha tan 4 alpha + tan 4 alpha tan alpha =

If alpha=(2 pi)/(7), then tan alpha tan2 alpha+tan2 alpha tan4 alpha+tan4 alpha tan alpha

" If "alpha=(2 pi)/(7)" ,and "tan alpha tan2 alpha+tan2 alpha tan4 alpha+tan4 alpha tan alpha=K" .Then "3k^(2)+4k+5=" ."

The number of solutions of the equation tan^(2)alpha-cos es^(2)alpha+4cot^(2)2 alpha+4tan alpha cot2 alpha=0in(0,(pi)/(2))-{(pi)/(4)} is

If alpha=(2 pi)/(7) then tan alpha tan^(7)2 alpha+tan2 alpha tan4 alpha+tan2 alpha tan4 alpha=

tan alpha+2tan2 alpha+4tan4 alpha+8cot8 alpha=cot alpha

If alpha is any real nubmer then: (sin^4alpha+sin^2alpha.cos^2alpha+cos^2alpha)/(cos^4alpha+sin^2alpha.cos^2alpha+sin^2alpha)=.... (a) 1 (b)2 (c) 3 (d) none of these

If alpha = (2pi)/7 prove that tan alpha *tan 2alpha + tan 2alpha* tan 4alpha + tan 4alpha* tan a = - 7

Prove that (sin alpha+sin2 alpha+sin3alpha+sin4alpha)/(cos alpha+cos 2 alpha+cos 3 alpha+cos4 alpha)=tan((5 alpha)/2)

tan alpha + 2 tan 2alpha + 4 tan 4 alpha + 8 cot 8 alpha =