Revision: Trigonometry Mathematics HSC Science Class 11 Tamil Nadu Board of Secondary Education

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

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

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

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

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

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

a cosθ + b sinθ = m  ......(i)

a sinθ - b cosθ = n  ......(ii)

Squaring and adding equations 1 and 2, we get,

(a cosθ + b sinθ)² + (a sinθ - b cosθ)² = m² + n²

⇒ a²cos²θ + b²sin²θ + 2ab sin θ cos θ + a²sin²θ + b²cos²θ - 2ab sin θ cos θ = m² + n²

⇒ a²cos²θ + b²sin²θ + a²sin²θ + b²cos²θ = m² + n²

⇒ a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ) = m² + n²

Using, sin²θ + cos²θ = 1

We get,

⇒ a² + b² = m² + n²

Theorem 1: For any real numbers x and y,                 
sin x = sin y implies x = nπ + (–1)n y, where n ∈ Z
Proof :If sin x = sin y, then
sin x – sin y = 0

 or  `2cos  (x+y)/2 sin  (x-y)/2= 0`

which gives `cos   (x+y)/2= 0` or `sin  (x-y)/2= 0`

Therefore `(x+y)/2= (2n+1) π/2` or `(x-y)/2= nπ`, where n ∈ Z

i.e. x = (2n + 1) π – y  or x = 2nπ + y, where n∈Z
Hence x = (2n + 1)π + (–1)2n + 1 y or x = 2nπ +(–1)2n y, where n ∈ Z.
Combining these two results, we get x = nπ + (–1)n y, where n ∈ Z.

Theorem 2: For any real numbers x and y, cos x = cos y, implies x = 2nπ ± y, where n ∈ Z
Proof: If cos x = cos y, then
cos x – cos y = 0   i.e.,

-2 sin `(x+y)/2 sin  (x-y)/2= 0`

Thus, `sin  (x+y)/2= 0` or `sin  (x-y)/2= 0`

Therefore, `(x+y)/2= nπ`  or `(x-y)/2= nπ`, where n ∈ Z
i.e. x = 2nπ – y  or x = 2nπ + y, where n ∈ Z Hence x = 2nπ ± y, where n ∈ Z

Theorem 3:Prove that if x and y are not odd mulitple of  `π/2`, then
tan x = tan y implies x = nπ + y, where n ∈ Z
Proof : If tan x = tan y,   then   tan x – tan y = 0
or `(sin x cos y- cos x sin y)/(cos x cos y)= 0`
which gives sin (x – y) = 0
Therefore x – y = nπ, i.e., x = nπ + y, where n ∈ Z.

LHS = `(cos A)/a + (cos B)/b + (cos C)/c`

`= ((("b"^2 + "c"^2 - "a"^2)/"2bc"))/"a" + ((("c"^2 + "a"^2 - "b"^2)/"2ca"))/"b" + ((("a"^2 + "b"^2 - "c"^2)/"2ab"))/"c"`

`= ("b"^2 + "c"^2 - "a"^2)/"2abc" + ("c"^2 + "a"^2 - "b"^2)/"2abc" + ("a"^2 + "b"^2 - "c"^2)/"2abc"`

`= ("b"^2 + "c"^2 - "a"^2 + "c"^2 + "a"^2 - "b"^2 + "a"^2 + "b"^2 - "c"^2)/"2abc"`

`= ("a"^2 + "b"^2 + "c"^2)/"2abc"`

= RHS

LHS = `(cos A)/a + (cos B)/b + (cos C)/c`

= `(b cos A + a cos B)/(ab) + (cos C)/c`

= `c/(ab) + (cos C)/c`    ...(By projection rule)

= `c/(ab) + (a^2 + b^2 - c^2)/(2 abc)`    ...(By cosine rule)

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

= `(a^2 + b^2 + c^2)/(2 abc)` = R.H.S.

Put x = cos θ

∴ θ = cos^–1 x

L.H.S. = `tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x)))`

= `tan^-1 ((sqrt(1 + cos θ) - sqrt(1 - cos θ))/(sqrt(1 + cos θ) + sqrt(1 - cos θ)))`

= `tan^-1 [(sqrt(2 cos^2(θ/2)) - sqrt(2 sin^2 (θ/2)))/(sqrt(2 cos^2 (θ/2)) + sqrt(2 sin^2 (θ/2)))]`

= `tan^-1 [(sqrt(2) cos (θ/2) - sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2) + sqrt(2) sin (θ/2))]`

= `tan^-1 [((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) - (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))/((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) + (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))]`

= `tan^-1 [(1 - tan(θ/2))/(1 + tan (θ/2))]`

= `tan^-1 [(tan  pi/4 - tan (θ/2))/(1 + tan  pi/4. tan (θ/2))]  ....[∵ tan  pi/4 =1]`

= `tan^-1 [tan (pi/4 - θ/2)]`

= `pi/4 - θ/2`

= `pi/4 - 1/2 cos^-1`x  .....[∵ θ = cos^–1 x]

∴ L.H.S. = R.H.S.

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.