Question

trigonometry identities

Answer 1

Hey There! This Is Your Answer ↓↓↓↓↓↓↓

• csc(x)=sin(x)1​

• \sin(x) = \dfrac{1}{\csc(x)}sin(x)=csc(x)1​

• \sec(x) = \dfrac{1}{\cos(x)}sec(x)=cos(x)1​

• \cos(x) = \dfrac{1}{\sec(x)}cos(x)=sec(x)1​

• \cot(x) = \dfrac{1}{\tan(x)} = \dfrac{\cos(x)}{\sin(x)}cot(x)=tan(x)1​=sin(x)cos(x)​

• \tan(x) = \dfrac{1}{\cot(x)} = \dfrac{\sin(x)}{\cos(x)}tan(x)=cot(x)1​=cos(x)sin(x)​

• sin2(t) + cos2(t) = 1

• tan2(t) + 1 = sec2(t)

• 1 + cot2(t) = csc2(t)


• sin(α + β) = sin(α) cos(β) + cos(α) sin(β)

• sin(α – β) = sin(α) cos(β) – cos(α) sin(β)

• cos(α + β) = cos(α) cos(β) – sin(α) sin(β)

• cos(α – β) = cos(α) cos(β) + sin(α) sin(β)

• \tan(\alpha + \beta) = \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)}tan(α+β)=1−tan(α)tan(β)tan(α)+tan(β)​

• \tan(\alpha - \beta) = \dfrac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha) \tan(\beta)}tan(α−β)=1+tan(α)tan(β)tan(α)−tan(β)​











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Answer 2

Answer:

Step-by-step explanation:

Sin²A + cos²A = 1

Sec²A. - Tan²A = 1

Cosec²A + cot²A = 1