Question #144324
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tan(θ)+cot(θ)=sec(θ)csc(θ)
1
Expert's answer
2020-11-18T19:35:53-0500

tan(θ)+cot(θ)=sin(θ)cos(θ)+cos(θ)sin(θ)=sin2(θ)+cos2(θ)cos(θ)sin(θ)=1cos(θ)sin(θ),sec(θ)csc(θ)=1cos(θ)1sin(θ)=1cos(θ)sin(θ).\tan(\theta)+\cot(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)} + \dfrac{\cos(\theta)}{\sin(\theta)} = \dfrac{\sin^2(\theta) + \cos^2(\theta)}{\cos(\theta)\sin(\theta)} = \dfrac{1}{\cos(\theta)\sin(\theta)}\,, \\ \sec(\theta)\csc(\theta) = \dfrac{1}{\cos(\theta)}\cdot \dfrac{1}{\sin(\theta)} = \dfrac{1}{\cos(\theta)\sin(\theta)}\,.

Therefore,

tan(θ)+cot(θ)=sec(θ)csc(θ).\tan(\theta)+\cot(\theta) = \sec(\theta)\csc(\theta) .



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