Formulae [1]
\[sineA=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\]
\[cosineA=\frac{\mathrm{Base}}{\text{Hypotenuse}}\]
\[tangentA=\frac{\text{Perpendicular}}{\mathrm{Base}}\]
\[cotangent A = \frac{\text{Base}}{\text{Perpendicular}}\]
\[secantA=\frac{\text{Hypotenuse}}{\mathrm{Base}}\]
\[cosecantA=\frac{\text{Hypotenuse}}{\text{Perpendicular}}\]
Theorems and Laws [5]
If `sin θ = 3/4` prove that `sqrt(("cosec"^2θ - cot^2θ)/(sec^2θ - 1)) = sqrt(7)/3`.
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
If `tan θ = a/b` then prove that `((a sin theta - b cos theta)/(a sin theta + b cos theta)) = ((a^2 - b^2))/((a^2 + b^2))`.
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)`
Prove that sin 75° – sin 15° = cos 105° + cos 15°
R.H.S = cos 105° + cos 15°
= cos(90° + 15°) + cos(90° – 75°)
= – sin 15° + sin 75°
= sin 75° – sin 15°
= L.H.S
If a cosθ + b sinθ = m and a sinθ - b cosθ = n, then show that a2 + b2 = m2 + n2
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.
Key Points
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.
