If latus recturn of the ellipse `x^2 tan^2 alpha+y^2 sec^2 alpha= 1` is `1/2` then `alpha(0 lt alpha lt pi)` is equal to

If the latus rectum of the ellipse x^(2)tan^(2)alpha+y^(2)sec^(2)alpha=1 is 1/2 then alpha=

If 0 lt alpha lt (pi)/(2) and the eccentricity of the ellipse x^(2) tan^(2) alpha + y^(2) sec^(2) alpha = 1 is 1//2 , then alpha

If 1+cos alpha+cos ^(2) alpha+….=2-sqrt(2) , thet alpha(0 lt alpha lt pi) is

Solve : tan ^(2) alpha + sec alpha - 1 = 0,0 le alpha lt 2pi.

If 0 lt alpha lt pi/2 , then sqrt(2cotalpha+1/(sin^2alpha)) is equal to

if (1+tan alpha )(1+tan4 alpha ) =2 where alpha in (0 , pi/16 ) then alpha equal to