Chapter 8 · Question 2

Show that tan⁡A=sin⁡Acos⁡A\tan A=\frac{\sin A}{\cos A}.

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Answer

Direct Answer

Since sin⁡A=oppositehypotenuse\sin A=\frac{\text{opposite}}{\text{hypotenuse}} and cos⁡A=adjacenthypotenuse\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}, sin⁡Acos⁡A=oppositeadjacent=tan⁡A\frac{\sin A}{\cos A}=\frac{\text{opposite}}{\text{adjacent}}=\tan A.

Simple Explanation

Divide sine by cosine; the hypotenuse cancels.

Exam-Ready Structure

Using definitions, sin⁡Acos⁡A=oppositehypotenuseadjacenthypotenuse=oppositeadjacent=tan⁡A\frac{\sin A}{\cos A}=\frac{\frac{\text{opposite}}{\text{hypotenuse}}}{\frac{\text{adjacent}}{\text{hypotenuse}}}=\frac{\text{opposite}}{\text{adjacent}}=\tan A.

Key Points

  • Since sin⁡A=oppositehypotenuse\sin A=\frac{\text{opposite}}{\text{hypotenuse}} and cos⁡A=adjacenthypotenuse\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}, sin⁡Acos⁡A=oppositeadjacent=tan⁡A\frac{\sin A}{\cos A}=\frac{\text{opposite}}{\text{adjacent}}=\tan A.