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

`cot theta = "𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒"/"𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒"`

Let x be the hypotenuse by applying Pythagoras theorem.

𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶²

𝑥² = 16 + 9

`x^2 = 25 => x = 5`

`sec theta = (AC)/(BC) = 5/3`

`cosec theta = (AC)/(AB) = 5/4`

On substituting in equation we get

`sqrt((sec theta - cosec theta)/(sec theta + cosec theta)) = sqrt((5/3 - 5/4)/(5/3 + 5/4))`

`= sqrt(((20 - 15)/12)/((20 + 15)/12)) = sqrt(5/35) = 1/sqrt7`

We have `sin theta = 3/4`

In ΔABC

`AC^2 = AB^2 + BC^2`

`=> (4)^2 = (3)^2 + BC^2`

`=> BC^2= 16 - 9`

`=> BC^2 = 7`

`=> BC = sqrt7`

`:. cosec theta = 4/3, sec theta = 4/sqrt7 and cot theta = sqrt7/3`

Now

L.H.S `sqrt((cosec^2 theta - cot^2 theta)/(sec^2 theta - 1))`

`= sqrt(((4/3)^2 - (sqrt7/3)^2)/((4/sqrt7)^2 - 1)`

`= sqrt((16/9 - 7/9)/(16/7 - 1)`

`=sqrt((9/9)/((16 - 7)/7 ))`

`= sqrt(7/9)`

`= sqrt7/3`

= R.H.S

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)`

LHS = sec 70° sin 70° + cos 20° cosec 70°

= sec (90° - 20°) sin 20° + cos 20° cosec (90° - 20°) 

`= "cosec" 20°. 1/("cosec" 20°)+ 1/(sec 20°)  sec 20°`

= 1 + 1

= 2 

= RHS 

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.