Question

all formula list of trignometry of class 11

Answer 1

Sin²x + Cos²x =
1

1+tan²x =
Sec²x

1+cot²x =
Cosec²x

sin(-x)
-Sin x

cos(-x)
cos x

sin(x+y) =
sin x cos y + cos x sin y

sin(a - b) =
sin a cos b - cos a sin b

cos(a +b) =
cos a cos b - sin a sinb

cos(a-b) =
cos a cos b + sin a sin b

In 4th Quadrant
cos & sec are +ve

In 2nd Quadrant
sin & cosec are +ve

In 1st Quadrant
All are +ve

In 3rd Quadrant
tan & cot are +ve

tan(a+b) =
(tan a + tan b) / (1 - tan a tan b)

tan(a-b) =
(tan a - tan b) / (1 + tan a tan b)

cot (α + β) =
(cot α cot β - 1) / (cot β + cot α)

cot(α - β) =
(cot α cot β +1) / (cot β - cot α)

sin 2θ
2sinθcosθ

sin2θ
2tanθ / ( 1 + tan²θ )

cos 2θ
cos²θ - sin²θ

cos 2θ
2cos²θ -1

cos 2θ
1 - 2sin²θ

cos 2θ
( 1 - tan²θ ) / ( 1 + tan²θ)

tan 2θ
2tanθ / 1-tan²θ

sin 3α
3sinα - 4sin³α

cos 3β
4cos³β - 3cosβ

tan 3θ
(3tanθ - tan³θ) / (1 - 3tan²θ)

cos α + cos β
2 cos[α+β/2] cos[α-β/2]

cosα - cosβ
−2 sin[α+β/2] sin[α−β/2]

sin α + sin β
2 sin [α+β/2] cos[α−β/2]

sin α − sinβ
2 cos[α+β/2] sin[α−β/2]

2 sin α cos β
sin(α+β) + sin(α−β)

2 cos α sin β
sin(α+β) - sin(α−β)

2 cos α cos β
cos(α+β) + cos(α−β)

2 sin α sin β
cos(α−β) - cos(α+β)

sin(α+β)sin(α−β)
sin²α − sin²β
cos²β − cos²α

cos(α+β)cos(α−β)
cos²α − sin²β
cos²β − sin²α

tan(α+β+γ)
(tanα + tanβ + tan γ − tanα tanβ tan γ) / (1- tan α tan β-tanβ tan γ - tan γ tanα)

-2 sin α sin β
cos(α+β) - cos(α−β)

{SubMultipleAngles}
sin A
2sin[A/2] cos[A/2] =
(2tan[A/2]) / ( 1 + tan²[A/2])

{SubMultipleAngles}
cos A
cos²[A/2] - sin²[A/2] =
2cos²[A/2] - 1 =
1 - 2sin²[A/2] =
(1-tan²[A/2]) / (1+tan²[A/2])

{SubMultipleAngles}
tan A
(2 tan [A/2]) / ( 1 − tan²[A/2] )

tan²A
( 1−cos A) / (1+cos A)

cos²A
( 1 + cos A) / 2

sin²A
( 1- cos A) / 2

Answer 2

Answer: all right make it as brainlist....

Step-by-step explanation: