Question

Can any one tell me the 3 formulas of trignmonetary chapter of class 10​

Answer 1

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$sin^</p><p>2A + cos^</p><p>2A = 1 \\ </p><p></p><p>tan^</p><p>2A + 1 = sec^</p><p> 2A \\ </p><p></p><p>cot^</p><p>2A + 1 = cosec^</p><p>2A</p><p></p><p>$

sin A =Perpendicular/Hypotenuse

cos A =Base/Hypotenuse

tan A= Perpendicular/Base

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Answer 2

Answer:

1 .sin A =Perpendicular/Hypotenuse

2.cos A= Base/Hypotenuse

3.tan A= Perpendicular/Base

4.cot A=Base/Perpendicular

5.cosec A= Hypotenuse/Perpendicular

6. sec A =Hypotenuse/Base

Reciprocal Relation Between Trigonometric Ratios

1 .tan A =sin A/cos A

2.cot A=cos A/sin A

3.cosec A= 1/sin A

4.sec A =1/cos A

Trigonometric Sign Functions

sin (-θ) = − sin θ

cos (−θ) = cos θ

tan (−θ) = − tan θ

cosec (−θ) = − cosec θ

sec (−θ) = sec θ

cot (−θ) = − cot θ

Trigonometric Identities

sin2A + cos2A = 1

tan2A + 1 = sec2A

cot2A + 1 = cosec2A

Periodic Identities

sin(2nπ + θ ) = sin θ

cos(2nπ + θ ) = cos θ

tan(2nπ + θ ) = tan θ

cot(2nπ + θ ) = cot θ

sec(2nπ + θ ) = sec θ

cosec(2nπ + θ ) = cosec θ

Complementary Ratios

Quadrant I

sin(π/2−θ) = cos θ

cos(π/2−θ) = sin θ

tan(π/2−θ) = cot θ

cot(π/2−θ) = tan θ

sec(π/2−θ) = cosec θ

cosec(π/2−θ) = sec θ

Quadrant II

sin(π−θ) = sin θ

cos(π−θ) = -cos θ

tan(π−θ) = -tan θ

cot(π−θ) = – cot θ

sec(π−θ) = -sec θ

cosec(π−θ) = cosec θ

Quadrant III

sin(π+ θ) = – sin θ

cos(π+ θ) = – cos θ

tan(π+ θ) = tan θ

cot(π+ θ) = cot θ

sec(π+ θ) = -sec θ

cosec(π+ θ) = -cosec θ

Quadrant IV

sin(2π− θ) = – sin θ

cos(2π− θ) = cos θ

tan(2π− θ) = – tan θ

cot(2π− θ) = – cot θ

sec(2π− θ) = sec θ

cosec(2π− θ) = -cosec θ

Sum and Difference of Two Angles

sin (A + B) = sin A cos B + cos A sin B

sin (A − B) = sin A cos B – cos A sin B

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

Double Angle Formulas

sin2A = 2sinA cosA = [2tan A + (1+tan2A)]

cos2A = cos2A–sin2A = 1–2sin2A = 2cos2A–1= [(1-tan2A)/(1+tan2A)]

tan 2A = (2 tan A)/(1-tan2A)

Thrice of Angle Formulas

sin3A = 3sinA – 4sin3A

cos3A = 4cos3A – 3cosA

tan3A = [3tanA–tan3A]/[1−3tan2A]

Step-by-step explanation:

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