Explained: "tan⁡α+cot⁡αsin⁡α−cos⁡α=sec2⁡α+csc2⁡α"

High School
Geometry
Trigonometry
Trigonometric Functions

Mylo O'Moore

Answered question

2021-06-07

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

Answer & Explanation

escumantsu

Skilled2021-06-08Added 98 answers

Work on the left side. Separate as: $\frac{\tan α + \cot α}{s i n α \times \cos α} = \frac{\tan α}{\sin α \times \cos α} + \frac{\cot α}{\sin α \times \cos α}$

Use the quotient identity for tangent and cotangent:

$\frac{\tan α + \cos α}{\sin α \times \cos α} = (\frac{\frac{\sin α}{\cos α}}{\sin α \times \cos α}) + (\frac{\frac{\cos α}{\sin α}}{\sin α \times \cos α})$

Simplify:

$(\frac{\tan α + \cot α}{\sin α \times \cos α}) = \frac{1}{\cos^{2}} α + \frac{1}{\sin^{2}} α$

Use the reciprocal identities for cosine and sine:

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

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