Question

Prove that:

`sec((3pi)/2 - theta) sec(theta - (5pi)/2) + tan((5pi)/2 + theta) tan(theta - (5pi)/2)` = - 1

Answer 1

`sec((3pi)/2 - theta)` = sec (270° – θ) = -cosec θ

[∵ For 270° – θ change T-ratio. So add ‘co’ infront ‘sec’, it becomes ‘cosec’]

`sec(theta - (5pi)/2) = sec (- ((5pi)/2 - theta))`

`= sec ((5pi)/2 - theta)`  ...[∵ sec(-θ) = θ]

= sec(450° – θ)

= sec (360° + (90° – θ))

= sec (90° – θ)

= cosec θ

[∵ For 90° – θ change in T-ratio. So add ‘co’ in front of ‘sec’ it becomes ‘cosec’]

`tan((5pi)/2 + theta)` = tan(450° + θ)

[∵ For 90° + θ, change in T-ratio. So add ‘co’ in front of ‘tan’ it becomes ‘cot’]

= tan (360° + (90° + θ))

= tan (90° + θ)

= -cot θ

`tan(theta - (5pi)/2) = tan(-((5pi)/2 - theta))`

`= - tan((5pi)/2 - theta)`  ...[∵ tan(-θ) = -tan θ]

= -tan(450° – θ)

= -tan(360° + (90° – θ))

= -tan(90° – θ)

= -cot θ

LHS = `sec((3pi)/2 - theta) sec(theta - (5pi)/2) + tan((5pi)/2 + theta) tan(theta - (5pi)/2)`

= -cosec θ (cosec θ) + (-cot θ) (-cot θ)

= -cosec² θ + cot² θ

= -(1 + cot² θ) + cot² θ [∵ 1 + cot² θ = cosec² θ]

= -1

= RHS