All Formulas for Class 10 NCERT math.
Answers
Answer:
Step-by-step explanation:
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