Question

If `(sec α)/("cosec"  β) = p` and `(tan α)/("cosec"  β) = q`, then prove that (p² – q²) sec²α = p².

Answer 1

P = `(sec α)/("cosec"  α)`

= `(1/(cos α))/(1/(sinα))`

= `(sin β)/(cos α)`

q = `(tan α)/("cosec"  β)`

= `(sin α)/(cos α) 1/(sin β)`

= `(sin α)/(cos α) xx (sin β)/1`

= `(sin α . sin β)/(cos α)`

Now, (p² – q²) sec² α

= `((sin^2 β)/(cos^2 α) - (sin^2 α sin^2 β)/(cos^2 α)) sec^2 α`

Take sin² β common from numerator

= `sin^2 β (1/(cos^2 α) - (sin^2 α)/(cos^2 α)) sec^2 α`

= sin² β (sec² α – tan² α) . sec² α

= `sin^2 β . 1/(cos^2 α)` (1) [∵ sec² α – tan² α = 1]

= `(sin^2 β)/(cos^2 α)`

= p²

∴ (p² – q²) sec² α = p²

Hence, proved.