Chapter 8 · Question 2

Show that tanA=sinAcosA\tan A=\frac{\sin A}{\cos A}.

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Answer

Direct Answer

Since sinA=oppositehypotenuse\sin A=\frac{\text{opposite}}{\text{hypotenuse}} and cosA=adjacenthypotenuse\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}, sinAcosA=oppositeadjacent=tanA\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, sinAcosA=oppositehypotenuseadjacenthypotenuse=oppositeadjacent=tanA\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 sinA=oppositehypotenuse\sin A=\frac{\text{opposite}}{\text{hypotenuse}} and cosA=adjacenthypotenuse\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}, sinAcosA=oppositeadjacent=tanA\frac{\sin A}{\cos A}=\frac{\text{opposite}}{\text{adjacent}}=\tan A.