Showing Trigonometric Identity. How do I demonstrate the next identity? (sin3θ/cosθ)+(cos3θ/sinθ)=2cot2θ?

Question

(sin3θ/cosθ)+(cos3θ/sinθ)=2cot2θ

(sin3θ/cosθ)+(cos3θ/sinθ)= 2cot2θ

Answer 1

Adding these fraction expressions (common denominator etc) gives
(sin3θsinθ + cos3θcosθ)/(cosθsinθ)

The top is the expansion of
cos(3θ - θ) which is cos2θ

The bottom is half of
sin2θ = 2sinθcosθ,
so multiply top and bottom by 2 and you have the result, since
cos2θ/sin2θ = cot2θ

Answer 2

2cot(2A) = 2/(tan(2A))

tan(2A) = (2tanA)/(1 - tanA^2)

2/(tan(2A)) = 2/((2tanA)/(1 - tanA^2))
2/(tan(2A)) = (2(1 - tanA^2))/(2tanA)
2/(tan(2A)) = (1 - tanA^2)/(tanA)

(1 - (sinA/cosA)^2)/(sinA/cosA)
((cosA^2 - sinA^2)/(cosA^2)) / (sinA/cosA)
((cosA^2 - sinA^2)/(cosA^2)) * (cosA/sinA)
((cosA)(cosA^2 - sinA^2))/((sinA)(cosA^2))
(cosA^2 - sinA^2)/(sinA * cosA)

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sin(3A) = sin(2A + A) = sin(2A)cosA + cos(2A)sinA
cos(3A) = cos(2A + A) = cos(2A)cosA - sin(2A)sinA

sin(2A) = 2sinAcosA
cos(2A) = 1 - 2sinA^2

sin(3A) = (2sinAcosA)cosA + (1 - 2sinA^2)sinA
cos(3A) = (1 - 2sinA^2)cosA - (2sinAcosA)sinA

sin(3A) = 2sinAcosA^2 + (1 - 2sinA^2)sinA
cos(3A) = (1 - 2sinA^2)cosA - 2cosAsinA^2

sin(3A) = (sinA)(2cosA^2 + 1 - 2sinA^2)
cos(3A) = (cosA)(1 - 2sinA^2 - 2sinA^2)

sin(3A) = (sinA)(2cosA^2 + 1 - 2(1 - cosA^2))
cos(3A) = (cosA)(1 - 4sinA^2)

sin(3A) = (sinA)(2cosA^2 + 1 - 2 + 2cosA^2) = (sinA)(4cosA^2 - 1)

(sin(3A)/cosA) + (cos(3A)/sinA) = ((sinA)(sin(3A)) + (cosA)(cos(3A))) / (sinA * cosA)

((sinA)((sinA)(4cosA^2 - 1)) + (cosA)((cosA)(1 - 4sinA^2))) / (sinA * cosA)

((sinA^2)(4cosA^2 - 1) + (cosA^2)(1 - 4sinA^2)) / (sinA * cosA)

(4sinA^2cosA^2 - sinA^2 + cosA^2 - 4sinA^2cosA^2) / (sinA * cosA)

(-sinA^2 + cosA^2) / (sinA * cosA)

(cosA^2 - sinA^2) / (sinsA * cosA)

so as you can see what i did in the first half and what i have done in the first have, you see that they are both equal, so therefore

(sin(3A)/cosA) + (cos(3A)/sinA) = 2cot(2A)

Sorry if was to long to understand, but i do hope you did understand it.

Answer 3

a million) a million+tan^2x=a million+sin^2x/cos^2x (cos^2x+sin^2x)/cos^2x (same deno.) all of us be attentive to that: sin^2x+cos^2x=a million then, a million+tan^2x=a million/cos^2x 2) a million/cos^2x= (sin^2x+cos^2x)/cos^2x a million+cos^2x =sin^2x/cos^2x+cos^2x/cos^2x with: sin^2x/cos^2x=tan^2x then, a million/cos^2x=a million+tan^2x