Question

If tan² θ = 1 – k², show that sec θ + tan³ θ cosec θ = (2 – k²)^3/2. Also, find the values of k for which this result holds

Answer 1

tan² θ = 1 – k²

1 + tan² θ = 1 + 1 – k²

sec²θ = (2 – k²)

sec²θ = `(2 – "k"^2)^(1/2` 

`sectheta + tan^3theta  "cosec"theta = 1/costheta + (sin^3theta)/(cos^3theta)*1/sintheta`

= `1/costheta + (sin^2theta)/(cos^3theta)`

= `1/costheta (1 + (sin^2theta)/(cos^2theta))`

= `sec theta (1 + tan^2 theta)`

= `sec theta * sec^2 theta`

= sec³θ

= `(2 - "k"^2)^(3/2)`

tan² θ = 1 – k²

When θ = `π/2`, tan `π/2` = ∞, not defined 2

When θ = 0, tan² 0 = 1 – k²

1 – k² = 0

⇒ k² = 1

⇒ k = ± 1

When θ = 45°, tan² 45° = 1 – k²

1 – k² = 1

⇒ – k² = 0

⇒ k = 0

When θ > 45°, say θ = 60°

tan² 60° = 1 – k²

= `(sqrt(3))^2`

= 1 – k²

3 = 1 – k² 

⇒ k² = 1 – 3

= – 2

∴ θ > 45°, k² is negative

⇒ k is imaginary

∴ k lies between – 1 and 1

⇒ k ∈ [– 1, 1]