Question

If cosec θ – sin θ = a³ and sec θ – cos θ = b³, then prove that a²b² (a² + b²) = 1

Answer 1

cosec θ – sin θ = a³ 

⇒ `1/sintheta - sintheta` = a³ 

(i.e) `(1 - sin^2theta)/sintheta` = a³ 

⇒ a³ = `(cos^2theta)/sintheta`

⇒ a = `((costheta)^(2/3))/((sintheta)^(1/3))`

Similary, sec θ – cos θ = b³ 

⇒ `1/costheta - costheta` = b³ 

`(1 - cos^2theta)/costheta` = b³ 

⇒ `(sin^2theta)/costheta` = b³ 

⇒ b = `((sintheta)^(2/3))/((costheta)^(1/3))`

Now a³ = `(cos^2theta)/sintheta`

b³ = `(sin^2theta)/costheta`

∴ a³b³ = `(cos^2theta)/sintheta * (sin^2theta)/costheta`

= sinθ cosθ
⇒ ab = `(sintheta costheta)^(1/3)`

= `sin^(2/3)theta cos^(2/3)theta`

To prove a²b²(a² + b²) = 1

a² + b² = `{[((costheta)^(2/3))/((sintheta)^(1/3))]^2 + [((sintheta)^(2/3))/((costheta)^(1/3))]^2}`

= `((costheta)^(4/3))/((sintheta)^(2/3)) + ((sintheta)^(4/3))/((costheta)^(2/3))`

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

= `1/((sinthetacostheta)^(2/3))`

L.H.S = a²b²(a² + b²)

= `(sintheta costheta)^(2/3) xx 1/((sintheta costheta)^(2/3))`

= 1

= R.H.S