Question

If tan²θ = 1 - a² , prove that secθ + tan³θcosecθ = ( 2 - a²)^3/2

Answer 1

Given, sec theta+tan^3 theta cosec theta can be written as

=root (1+tan^2theta) + tan^2 theta  * tan theta * root (1+cot^2 theta)

(Given tan^2 theta = 1 - a^2)

= root 1+(1-a^2)+(1-a^2) * root (1-a^2) * root (1+(1/tan^2 theta))  

= root (2-a^2) + (1-a^2) * root (1-a^2) * root (1+1/(1-a^2))

= root(2-a^2)+(1-a^2) * root (2-a^2)

=root (2-a^2) * (1+1-a^2)

=root (2-a^2) * (2-a^2)

=(2-a^2)^1/3

=(2-a^2)^3/2


Hope this helps!