Solved: "tan⁡α+cot⁡αsin⁡α−cos⁡α=sec2α+csc2α"

High School
Geometry
Trigonometry
Trigonometric Functions

Alyce Wilkinson

Answered question

2021-09-04

$\frac{\tan α + \cot α}{\sin α - \cos α} = \sec^{2} α + \csc^{2} α$

Answer & Explanation

Benedict

Skilled2021-09-05Added 108 answers

Work on the left side. Separate as:

Work on the left side.
Separate as:
$\frac{\tan α + \cot α}{\sin α \cdot \cos α} = (\frac{\tan}{α} \sin α \cdot \cos α + (\frac{\cot}{α} \sin α \cdot \cos α))$
Use the quotient identity for tangent and cotangent:
$\frac{\tan α + \cos α}{\sin α \cdot \cos α} = ((\frac{\frac{\sin}{α} \cos α}{\sin α \cdot \cos α}) + (\frac{\frac{\cos}{α} \sin α}{\sin α \cdot \cos α}))$
Simplify:
$(\frac{\tan α + \cot α}{\sin α} \cdot \cos α) = \frac{1}{\cos^{2} α} + \frac{1}{\sin^{2} α}$
Use the reciprocal identities for cosine and sine:
$\frac{\tan α + \cot α}{\sin α \cdot \cos α} = \sec^{2} α + \csc^{2} α$

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