Question

Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`

Answer 1

Proof: L.H.S. = `tan"A"/(1 + tan^2 "A")^2 + cot"A"/(1 + cot^2 "A")^2`

= `tan "A"/(sec^2"A")^2 + cot "A"/("cosec"^2"A")^2`  ......`[(∵ 1 + cot^2θ = "cosec"^2θ),(1 + tan^2θ = sec^2θ)]`

= `tan "A"/sec^4"A" + cot "A"/("cosec"^4"A")`

= `sin "A"/cos "A" xx 1/(sec^4 "A") + cos "A"/sin "A" xx 1/("cosec"^4 "A")`

= `sin "A"/cos "A" xx cos^4"A" + cos "A"/sin "A" xx sin^4"A"`

= sinA × cos³A + cosA × sin³A

= sinA cosA (cos²A + sin²A)

= sinA cosA  (1) ......[∵ cos²A + sin²A = 1]

= sinA.cosA

= R.H.S

L.H.S. = R.H.S.

Hence proved.