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

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

For the hyperbola x^2/ cos^2 alpha - y^2 /sin^2 alpha = 1;(0 lt alphalt pi/4) . Which of the following remains constant when alpha varies?

If sin (alpha + beta) = 4/5 , cos (alpha - beta) = (12)/(13) 0 lt alpha lt (pi)/(4), 0 lt beta lt (pi)/(4), find tan 2 alpha .

If sin alpha +cos alpha =(sqrt(3)+1)/(2), 0 lt alpha lt 2pi , then possible values "tan"(alpha)/(2) can take is/are :

If the equation k sin x + sqrt(k-2)cos x+(tan alpha+cot alpha)=0, 0 lt alpha lt(pi)/(2) , possesses real solution, then k belongs to

The sides of the triangle are sin alpha , cos alpha and sqrt(1+sinalphacosalpha) for some 0 lt alpha lt (pi)/2 . Then the greatest angle of the triangle is

If a sec alpha-c tan alpha=d and b sec alpha + d tan alpha=c , then eliminate alpha from above equations.

If Delta_(1) is the area of the triangle with vertices (0, 0), (a tan alpha, b cot alpha), (a sin alpha, b cos alpha), Delta_(2) is the area of the triangle with vertices (a sec^(2) alpha, b cos ec^(2) alpha), (a + a sin^(2)alpha, b + b cos^(2)alpha) and Delta_(3) is the area of the triangle with vertices (0,0), (a tan alpha, -b cot alpha), (a sin alpha, b cos alpha) . Show that there is no value of alpha for which Delta_(1), Delta_(2) and Delta_(3) are in GP.

If sec (alpha - 2 beta), sec alpha , sec (alpha + 2 beta) are in arithmetical progressin then cos ^(2) alpha = lamda cos ^(2) beta (beta ne n pi, n in I) the value of lamda is:

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