3. `sec^2theta - 1 =?`

Quadratic equation 8 "sec"^(2) theta - 6 "sec" theta + 1 = 0 has

If a chord joining P(a sec theta, a tan theta), Q(a sec alpha, a tan alpha) on the hyperbola x^(2)-y^(2) =a^(2) is the normal at P, then tan alpha is (a) tan theta (4 sec^(2) theta+1) (b) tan theta (4 sec^(2) theta -1) (c) tan theta (2 sec^(2) theta -1) (d) tan theta (1-2 sec^(2) theta)

If the length of the latusrectum of the ellipse x^(2) tan^(2) theta + y^(2) sec^(2) theta = 1 is 1/2, then theta =

value of 1+tan^2theta options are 1] cot theta 2] cosec theta 3] sec^2theta 4] tan^2theta

prove that : (tan 2^n theta)/(tan theta) = (1+sec 2theta) (1+ sec 2^2 theta) (1+sec 2^3 theta) … (1+ sec 2^n theta)

Prove that (sec^(2)theta-1)/(tan^(2)theta)=1

Prove that sec^(2)theta-1-tan^(2)theta=0

Prove: (sec^2theta-1)(cos e c^2theta-1)=1

If sec^2 theta (1 + sin theta) (1 — sin theta) = k , then find the value of k.

If the normal at 'theta' on the hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 meets the transverse axis at G , and A and A' are the vertices of the hyperbola , then AC.A'G= (a) a^2(e^4 sec^2 theta-1) (b) a^2(e^4 tan^2 theta-1) (c) b^2(e^4 sec^2 theta-1) (d) b^2(e^4 sec^2 theta+1)