Chapter 8 · Question 1

Define the six trigonometric ratios for an acute angle $A$ in a right triangle.

Answer Revealed

Direct Answer:

For angle

A

\sin A=\frac{\text{opposite}}{\text{hypotenuse}}

\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}

\tan A=\frac{\text{opposite}}{\text{adjacent}}

\cosec A=\frac1{\sin A}

\sec A=\frac1{\cos A}

\cot A=\frac1{\tan A}

Sine, cosine, and tangent compare sides of a right triangle. The other three are reciprocals.

In a right triangle, choose acute angle

A

. The side opposite

A

, the side adjacent to

A

, and the hypotenuse determine the ratios. Thus

\sin A

\cos A

, and

\tan A

are side ratios, while

\cosec A

\sec A

, and

\cot A

are their reciprocals.

For angle $A$ : $\sin A=\frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan A=\frac{\text{opposite}}{\text{adjacent}}$ , $\cosec A=\frac1{\sin A}$ , $\sec A=\frac1{\cos A}$ , $\cot A=\frac1{\tan A}$ .
Use the NCERT formula or theorem carefully.
Write units and final conclusion where applicable.

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