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Question
Show that cot(A + 15°) – tan(A – 15°) = `(4cos2"A")/(1 + 2 sin2"A")`
Solution
L.H.S = `(cos("A" + 15^circ))/(sin("A" + 15^circ)) - (sin("A" - 15^circ))/(cos("A" - 15^circ))`
= `(cos("A" + 15^circ)cos("A" - 15^circ) - sin("A" + 15^circ)sin("A" - 15^circ))/(sin("A" + 15^circ) cos("A" - 15^circ))`
= `(cos("A" + 15^circ + "A" - 15^circ))/(1/2[sin("A" + 15^circ + "A" - 15^circ) + sin("A" + 15^circ - "A" + 15^circ)]`
= `(2cos2"A")/(sin2"A" + sin30^circ)`
= `(2cos2"A")/(1/2 + sin2"A")`
= `(4cos2"A")/(1 + 2sin 2"A")`
= R.H.S
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