Prove that cos(3π/2+x)cos(2π+x) [cot(3π/2-x) + cot(2π + x)] = 1

Question

Prove that cos(3π/2+x)cos(2π+x) [cot(3π/2-x) + cot(2π + x)] = 1

Prove that cos(3π/2+x)cos(2π+x) [cot(3π/2-x) + cot(2π + x)] = 1

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Using the sum of angles identities:
cos(3π/2+x) = cos(3π/2)cos(x) - sin(3π/2)sin(x) = sin(x)
cos(2π+x) = cos(x)
cot(3π/2-x) = [cot(3π/2)cot(x) + 1]/[cot(x) - cot(3π/2) = 1/cot(x) = tan(x)
cot(2π+x) = cot(x)

Substituting into the original equation:
cos(3π/2+x)cos(2π+x) [cot(3π/2-x) + cot(2π + x)] = 1
sin(x)cos(x)[tan(x)+cot(x)] = 1
sin(x)cos(x)[sin(x)/cos(x) + cos(x)/sin(x)] = 1
sin²x + cos²x = 1