Question

(tanα+cotα)/(sinα-cosα)=sec^2α+csc^2α

Trigonometric Functions
ANSWERED
asked 2021-06-07

\(\frac{\tan\alpha+\cot\alpha}{\sin\alpha-\cos\alpha}=\sec^2\alpha+\csc^2\alpha\)

Answers (1)

2021-06-08

Work on the left side. Separate as:  \(\frac{\tan\alpha+\cot\alpha}{sinα\times\cos\alpha}=\frac{\tan\alpha}{\sin\alpha\times\cos\alpha}+\frac{\cot\alpha}{\sin\alpha\times\cos\alpha}\)

Use the quotient identity for tangent and cotangent:

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

Simplify:

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

Use the reciprocal identities for cosine and sine:

\(\frac{\tan\alpha+\cot\alpha}{\sin\alpha\times\cos\alpha}=\sec^2\alpha+\csc^2\alpha\)

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