Answer to Question #144324 in Trigonometry for Shahir Sheikh

Question

Please verify this identity on your own paper and upload a picture of your work.

tan(θ)+cot(θ)=sec(θ)csc(θ)

Answer 1

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

Therefore,

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