Question

Prove the following:

`(2cos2"A" + 1)/(2cos2"A" - 1)` = tan(60° + A) tan(60° − A)

Answer 1

R.H.S. = tan(60° + A) tan(60° − A)

= `(sin(60^circ + "A")sin(60^circ - "A"))/(cos(60^circ + "A")cos(60^circ - "A")`

= `(2sin(60^circ + "A")sin(60^circ - "A"))/(2cos(60^circ + "A")cos(60^circ - "A")`

= `(cos[60^circ + "A" - (60^circ - "A")] - cos(60^circ + "A" + 60^circ - "A"))/(cos(60^circ + "A" + 60^circ - "A") + cos[60^circ + "A" - (60^circ - "A")]`

= `(cos2"A" - cos120^circ)/(cos120^circ - cos2"A")`

= `(cos2"A" - cos(180^circ - 60^circ))/(cos(180^circ - 60^circ) + cos2"A")`

= `(cos2"A" - (- cos 60^circ))/(- cos60^circ + cos2"A")`

= `(cos2"A" + 1/2)/(-1/2 + cos2"A")`

= `(2cos2"A" + 1)/(2cos2"A" - 1)`

= L.H.S.