Question

cot attores of
coto
cosec at
It cosa
Sind​

Answer 1

Answer:

a right-angled triangle,

Sinθ= Opposite Side/Hypotenuse

Cosθ= Adjacent Side/Hypotenuse

Tanθ= Sinθ/Cosθ  = Opposite Side/Adjacent Side

Cosecθ = 1/Sinθ= Hypotenuse/Opposite Side

Secθ = 1/Cosθ = Hypotenuse/Adjacent Side

Cotθ = 1/tanθ = Cosθ/Sinθ = Adjacent Side/Opposite Side

SinθCosecθ = CosθSecθ = TanθCotθ = 1

Sin(90-θ) = Cosθ, Cos(90-θ) = Sinθ

Sin²θ + Cos²θ = 1

Tan²θ + 1 = Sec²θ

Cot²θ + 1 = Cosec²θ

Addition and subtraction formula:-

Sin(A+B) = SinACosB + CosASinB

Sin(A-B) = SinACosb - CosASinB

Cos(A+B) = CosACosB - SinASinB

Cos(A-B) = CosACosB + SinASinB

Tan(A+B) = (TanA+TanB)/(1-TanATanB)

Tan(A-B) = (TanA - TanB)/(1+TanATanB)

Cot (A+B) = (CotACotB-1)/(CotA + CotB)

Cot(A-B) = (CotACotB+1)/(CotB-CotA)

Sin(A+B)+Sin(A-B) = 2SinACosB

Sin(A+B)-Sin(A-B) = 2CosASinB

Cos(A+B)+Cos(A-B) = 2CosACosB

Edited : Cos(A - B) - Cos(A + B) = 2SinASinB

SinC + SinD = 2Sin[(C+D)/2]Cos[(C-D)/2]

SinC - SinD = 2Cos[(C+D)/2]Sin[(C-D)/2]

CosC + CosD = 2Cos[(C+D)/2]Cos[(C-D)/2]

CosC - CosD = 2Sin[(C+D)/2]Sin[(D-C)/2]

Sin2θ = 2SinθCosθ = (2tanθ)/(1+tan²θ)

Cos2θ = Cos²θ - Sin²θ = 2Cos²θ - 1= 1 - 2Sin²θ =

(1-tan²θ)/(1+tan²θ)

Tan2θ = 2tan θ/(1-tan²θ)

HOPE you understand and like