Question

If tan^2A= (1-e^2), show that secA+tan^3A= (2-e^2)^3/2. ​

Answer 1

Answer:

Step-by-step explanation:

tan²θ=1-a²

LHS

=secθ+tan³θcosecθ

=√(1+tan²θ)+tan²θ×tanθ×√(1+cot²θ)

[∵, sec²θ-tan²θ=1 and cosec²θ-cot²θ=1]

=√1+(1-a²)+(1-a²)×√(1-a²)×√{1+(1/tan²θ)}

=√(2-a²)+(1-a²)×√(1-a²)×√{1+1/(1-a²)}

=√(2-a²)+(1-a²)×√(1-a²)×√{(1-a²+1)/(1-a²)}

=√(2-a²)+(1-a²)×√(2-a²)

=√(2-a²)×(1+1-a²)

=√(2-a²)×(2-a²)

=(2-a²)¹/²⁺¹

=(2-a²)³/²

=RHS (Proved)