Revision: Trigonometry Mathematics SSLC (English Medium) Class 9 Tamil Nadu Board of Secondary Education

\[sineA=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\]

\[cosineA=\frac{\mathrm{Base}}{\text{Hypotenuse}}\]

\[tangentA=\frac{\text{Perpendicular}}{\mathrm{Base}}\]

\[cotangent A = \frac{\text{Base}}{\text{Perpendicular}}\]

\[secantA=\frac{\text{Hypotenuse}}{\mathrm{Base}}\]

\[cosecantA=\frac{\text{Hypotenuse}}{\text{Perpendicular}}\]

For an acute angle A, 

1. sin (90° - A) = cos A
2. cos (90° - A) = sin A
3. tan (90° - A) = cot A
4. cot (90° - A) = tan A
5. sec (90° - A) = cosec A
6. cosec (90° - A) = sec A

Let `(a sin  theta - b cos theta)/(a sin theta + b cos theta)`

Divide both Nr and Dr with cos θ of (a)

`((a sin theta - b cos theta)/cos theta)/((a sin theta + b cos theta)/cos theta)`

`= (tan theta - b)/(a tan theta + b)`

`=(a xx (a/b) - b)/(a xx (a/b) + b)`

`= (a^2 - b^2)/(a^2 + b^2)`

∵ tan A = cot B

tan A = tan (90° – B)

A = 90° – B

A + B = 90°. Proved

R.H.S. = `(sec^2A)/("cosec"^2A)`

= `(1 + tan^2A)/(1 + cot^2A)`   ...`[(∵ 1 + tan^2A = sec^2A),(1 + cot^2A = "cosec"^2A)]`

= `(1 + (sin^2A)/(cos^2A))/(1 + (cos^2A)/(sin^2A))`

= `((cos^2A  +  sin^2A)/(cos^2A))/((sin^2A  +  cos^2A)/(sin^2A))`

= `(1/(cos^2A))/(1/(sin^2A))`   ...[∵ sin²A + cos²A = 1]

= `(sin^2A)/(cos^2A)`

= tan²A

= tan A . tan A

= `(tan A)/(cot A)`

= L.H.S.

∴ `(tan A)/(cot A) = (sec^2A)/("cosec"^2A)`

For an acute angle A in a right-angled triangle:

• Hypotenuse is the side opposite the right angle.

• Perpendicular is the side opposite angle A.

• Base is the side adjacent to angle A.