Trigonometric Ratios
In right angled triangle, there are three sides which have there own special names: hypotenuse, perpendicular and base.
1. The side opposite to 90° is called hypotenuse(h).
2. The side opposite to reference angle is called perpendicular(p).
3. The side in between reference angle and 90° is called base(b).
By using these three sides, we can form six ratios. These ratios are called Trigonometric ratios for the reference angle θ. Let us define them.
i) sin θ = perpendicular/hypotenuse = p/h
ii) cos θ = base/hypotenuse = b/h
iii) tan θ = perpendicular/base = p/b
iv) cosec θ = hypotenuse/perpendicular = h/p
v) sec θ = hypotenuse/base = h/b
vi) cot θ = base/perpendicular = b/p
Important Formula
i) sin2θ + cos2θ = 1
ii) sin2θ = 1 - cos2θ
iii) cos2θ = 1 - sin2θ
iv) sinθ = (√1 - cos2θ)
v) cosθ = (√1-sin2θ)
vi) sec2θ -tan2θ = 1
vii) sec2θ = 1 + tan2θ
viii) tan2θ = sec2θ - 1
ix) secθ = (√1 + tan2θ)
x) tanθ = (√sec2θ - 1)
xi) cosec2θ - cot2θ = 1
xii) cosec2θ = 1 + cot2θ
xiii) cot2θ = cosec2θ - 1
xiv) cosecθ = (√1 + cot2θ)
xv) cotθ = (√cosec2θ - 1)
xvi) tanθ = sinθ/cosθ
xvi) cotθ = cosθ/sinθ
[NOTE: sinθ = 1/cosecθ , cosθ = 1/secθ , tanθ = 1/cotθ , cotθ = 1/tanθ , secθ = 1/cosθ , cosecθ = 1/sinθ]
Ex-5.4
1. Simplify
a) -sinA(cosA - sinA) - cosA(sinA - cosA)
Ans: Here,
= -sinA.cosA + sin2A - cosA.sinA + cos2A
= -2sinA.cosA + 1 [sin2A + cos2A = 1]
= 1 - 2sinA.cosA
b) tanA(secA + 1) + secA(1 - tanA)
Ans: Here,
= tanA.secA + tanA + secA - secA.tanA
= tanA.secA + tanA + secA - tanA.secA
= tanA + secA
c) (2tanA + 1) (3tanA - 8)
Ans: Here,
= 2tanA(3tanA - 8) + 1(3tanA - 8)
= 6tan2A - 16tanA + 3tanA - 8
= 6tan2A -13tanA -8
d) (sinB - cosB) (sinB + cosB)
Ans: Here,
= sinB(sinB + cosB) - cosB(sinB + cosB)
= sin2B + sinB.cosB - sinB.cosB - cos2B
= sin2B + sinB.cosB - sinB.cosB - cos2B
= sin2B - cos2B
e) (1 - secA) (1 + secA) (1 + sec2A)
Ans: Here,
= (1 - sec2A) (1 + sec2A) [(a-b)(a+b)= a2-b2]
= 1(1 + sec2A) - sec2A(1 + sec2A)
= 1 + sec2A - sec2A - sec4A
= 1 - sec4A
f) (5 - cosecβ) (2cosecβ + 4)
Ans: Here,
= 5(2cosecβ + 4) - cosecβ(2cosecβ + 4)
= 10cosecβ + 20 - 2cosec2β - 4cosecβ
= 20 + 6cosecβ - 2cosec2β
g) (sinA - cosA) (sin2A + sinA.cosA + cos2A)
Ans: Here,
= sin3A - cos3A [a3-b3=(a-b)(a2+ab+b2)]
h) (1 + tanA) (1 - tanA + tan2A)
Ans: Here,
= 1 + tan3A [a3+b3=(a+b)(a2-ab+b2)]
i) (2tanC - cosecB) (3tanC + 2cosecB)
Ans: Here,
= 2tanC(3tanC+2cosecB)-cosecB(3tanC+2cosecB)
= 6tan2C+4tanC.cosecB-3tanC.cosecB-2cosec2B
= 6tan2C+tanC.cosecB-2cosec2B
j) sin2x (sin2x- 1) - 3(sin4x - 1)
Ans: Here,
= sin4x - sin2x - 3sin4x + 9
= -2sin4x - sin2x + 9
3. Prove the following identities.
4. Show that:
5. Prove that:
