If sin π 4 cot θ = cos π 4 tan θ , then θ =

If sinπ4cotθ=cosπ4tanθ, then θ=
  1. 2nπ+π4
  2. 2nπ±π4
  3. 2nπ-π4
  4. nπ+π4

Solution

Given sinπ4cotθ=cosπ4tanθ

Using sinπ2-θ=cosθ,

sinπ4cotθ=sinπ2-π4tanθ

π4cotθ=π2-π4tanθ

π4tanθ+cotθ=π2

tanθ+cotθ=2

sinθcosθ+cosθsinθ=2

sin2θ+cos2θcosθsinθ=2

1sinθcosθ=2

2sinθcosθ=1

Using 2sinθcosθ=sin2θ,

sin2θ=1

The solution of sinx=1 is x=2nπ+π2, nZ.

2θ=2nπ+π2, nZ

θ=nπ+π4.

Hence, the solution of the given equation is nπ+π4.

Asked in: AP EAMCET 2021 (20 Aug Shift 1)

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