hello frnds...⭐
class 10
all trignometry formulas.....
note:-15 points.....
Answers
(perpendicular )2 + ( base )2 = ( hypotenuse )2
SinA = P / H
CosA = B / H
TanA = P / B
CotA = B / P
CosecA = H / P
SecA = H/B
Sin2A + Cos2A = 1
Tan2A + 1 = Sec2A
Cot2A + 1 = Cosec2A
TanA = SinA / CosA
2. CotA = CosA / SinA
3. CosecA = 1 / SinA
4. SecA = 1 / CosA
Sin (A +B) =SinA . CosB + CosA . SinB
Sin (A – B) = SinA . CosB – CosA . SinB
Cos (A + B) = CosA . CosB – SinA . SinB
Cos (A – B) = CosA. CosB + SinA . SinB
Tan (A + B) = TanA + TanB / 1 – TanA . TanB
Tan (A – B) = TanA –TanB / 1 + TanA . TanB
Sin ( A + B) . Sin (A – B) = Sin2A – Sin2B = Cos2B – Cos2A
Cos (A + B) . Cos (A – B) = Cos2A – Sin2B = Cos2B – Sin2A
Sin2A = 2 . SinA . CosA = 2 . TanA / (1 + Tan2A)
Cos2A = Cos2A – Sin2A = 1 – 2Sin2A = 2Cos2A – 1 = (1 – Tan2A) / (1 + Tan2A)
Tan2A = 2TanA / (1 – Tan2A)
Sin3A = 3 . SinA – 4 . Sin3A
Cos3A = 4 . Cos3A – 3 . CosA
Tan3A = (3TanA – Tan3A) / (1 – 3Tan2A)
SinA + SinB = 2 Sin (A + B)/2 Cos (A – B)/2
SinA – SinB = 2 Sin (A – B)/2 Cos (A + B)/2
CosA + CosB = 2 Cos(A – B)/2 Cos (A + B)/2
CosA – CosB = 2 Sin(B – A)/2 Sin (A + B)/2
TanA + TanB = Sin (A + B) / CosA . CosB
SinA CosB = Sin (A + B) + Sin (A – B)
CosA SinB = Sin (A + B) – Sin (A – B)
CosA CosB = Cos (A + B) + Cos (A – B)
SinA SinB = Cos (A – B) – Cos (A + B)
I hope this will help you
XD
The reciprocal relationship between different Trigonometric Functions are as under:
tan
θ
=
1
cot
θ
=
sin
θ
cos
θ
cot
θ
=
1
tan
θ
=
cos
θ
sin
θ
c
o
s
e
c
θ
=
1
sin
θ
sec
θ
=
1
cos
θ