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$\mathsf {If \: {tan}^{2} \theta \: = \: 1 \: - \: {e}^{2}, \: then \: sec \theta \: + \: {tan}^{3} \theta.cosec \theta = }$

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Answer 1

Answer:

(2- e²)^3/2

Step-by-step explanation:

Given Equation is tan²θ = 1 - e²

It can be written as:

⇒ 1 + tan²θ = 1 + 1 - e²

⇒ sec²θ = 2 - e² {∴ sec²θ - tan²θ = 1}

Now,

= secθ + tan³θ * cosecθ

= (1/cosθ) + (sin³θ/cos³θ) * (1/sinθ)

= (1/cosθ) + (sin²θ/cos³θ)

= (1/cosθ) + (1-cos²θ/cos³θ)

= (1/cosθ) + (1/cos³θ) - (cos²θ/cos³θ)

= (1/cosθ) + (1/cos³θ) - (1/cosθ)

= (1/cos³θ)

= sec³θ

= (2 - e²)^3/2

Hope it helps!

Answer 2

Step-by-step explanation:

tan²θ=1-a²

=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