Question 352159: Prove that cos(x)-sin(x)cot(x)=0 and
tan^5(x)=tan^3(x)sec^3(x)-tan^3(x)
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! Prove that cos(x)-sin(x)cot(x)=0 and
tan^5(x)=tan^3(x)sec^3(x)-tan^3(x)
sohcahtoa
sin = opp/hyp, cos = adj/hyp, tan = opp/adj
adj, opp, hyp sides of a right triangle
tan = sin/cos, cot = 1/tan = cos/sin
sec = 1/cos, csc = 1/sin
sin^2 + cos^2 = 1 (Pythagorean Identity)
sin^2/cos^2 + 1 = 1/cos^2 = tan^2 + 1 = sec^2 (Pythagorean Identity)
1 + cos^2/sin^2 = 1/sin^2 = 1 + cot^2 = csc^2 (Pythagorean Identity)
cos(x) - sin(x)cot(x) = 0, prove
cos(x) - sin(x) * cos(x)/sin(x) = 0, prove (replaced cot(x))
cos(x) - cos(x) = 0, true
tan^5(x) = tan^3(x)sec^3(x) - tan^3(x), prove
tan^5(x) = tan^3(x)(sec^3(x) - 1), prove
sec^2 - 1 = tan^2, NOT sec^3 - 1, the exponent on the sec^3(x) is wrong
the exponent on the sec^3(x) should be a 2 making it sec^2(x)
when corrected:
tan^5(x) = tan^3(x)(sec^2(x) - 1), prove
tan^5(x) = tan^3(x)(tan^2(x)), prove
tan^5(x) = tan^5(x), true
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