Question

If tan 2 theta = 1 - a 2 , prove that (sec theta + tan 3 theta cosec theta) = (2-a 2 ) 3/2

Answer 1

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)

Answer 2

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

( HENCE PROVED)