Question

Important formulas of Trigonometry for class 10​

Answer 1

$\huge\red{\boxed{\sf AnSwer }}$

• sin A → Perpendicular/Hypotenuse
• cos A→ Base/Hypotenuse
• tan A→ Perpendicular/Base
• cos A→ Base/Perpendicular
• cosec A→ Hypotenuse/Perpendicular
• sec A→ Hypotenuse/Base

Reciprocal Relation Between Trigonometric Ratios

• tan A →sin A/cos A
• cot A→ cos A/sin A
• cosec A→ 1/sin A
• 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)]

Answer 2

Answer:

hi

good afternoon

have a nice day