Question

All Formulas for Class 10 NCERT math.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Trigonometry:

In a right-angled triangle, the Pythagoras theorem states

(perpendicular )2 + ( base )2 = ( hypotenuse )2

Important trigonometric properties: (with P = perpendicular, B = base and H = hypotenuse)

SinA = P / H

CosA = B / H

TanA = P / B

CotA = B / P

CosecA = H / P

SecA = H/B

Trigonometric Identities:

sin2A + cos2A=1

tan2A +1 = sec2A

cot2A + 1= cosec2A

Relations between trigonometric identities are given below:

Trigonometric Ratios of Complementary Angles are given as follows:

sin (90° – A) = cos A

cos (90° – A) = sin A

tan (90° – A) = cot A

cot (90° – A) = tan A

sec (90° – A) = cosec A

cosec (90° – A) = sec A

(i) (a + b)2 = a2 + 2ab + b2

(ii) (a – b)2 = a2 – 2ab + b2

(iii) a2 – b2 = (a + b) (a – b)

(iv) (a + b)3 = a3 + b3 + 3ab(a + b)

(v) (a – b)3 = a3 – b3 – 3ab(a – b)

(vi) a3 + b3 = (a + b) (a2 – ab + b2)

(vii) a3 – b3 = (a – b) (a2 + ab + b2)

(viii) a4 – b4 = (a2)2 – (b2)2 = (a2 + b2) (a2 – b2) = (a2 + b2) (a + b) (a – b)

(ix) (a + b + c) 2 = a2 + b2 + c2 + 2ab + 2bc + 2ac

(x) (a + b – c) 2 = a2 + b2 + c2 + 2ab – 2bc – 2ca

(xi) (a – b + c)2 = a2 + b2 + c2 – 2ab – 2bc + 2ca

(xii) (a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca

(xiii) a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca