trigonometric identities cot

1. sin A = cos (90 o - A). 2. sin A / cos a = tan A. 3. tan A x tan (90 o - A) = 1. 4. sin 2 A + cos 2 A = 1. 5. 1 + tan 2 A = sec 2 A. 6. 1 + cot 2 A = cosec 2 A. Where 'A' is the angl..

1. sin A = cos (90 o - A) 2. 3. tan A x tan (90 o - A) = 1 4. sin 2 A + cos 2 A = 1 5. 1 + tan 2 A = sec 2 A 6. 1 + cot 2 A = cosec 2 A Let us prove the above identities. Let D ABC be a right-angled triangle with B = 90 o . Let BC = a, AC = b and AB = c. (1..

The trigonometric ratio's are: Sine, Cosine, Tangent, Cotangent, Secant, and Cosecan..

In the above topics many identities have been discussed. They are true for all values of the angles for which trigonometric functions are defined. In this section we prove identities, where a certain relationship exists among the angles considered. Many int..

When the integrand consists of trigonometric function, we use suitable trigonometric identities to simplify the function so that it can be integrated. Few identities are given below for ready reference. (1) (2) (3) (4) (5) (7) (..

Introduction - Let us recapitulate the trigonometric ratios (t-ratios). There are six t-ratios. D ABC is a right-angled triangle, B = 90 o . (i) In short, (ii) In shor..

The circle whose radius is 1 unit whose centre is the origin of a rectangular co-ordinate system is called the unit circle. 1. cos q = x. 2. sin q = y. 3. tan q = y/x. 4. sec q = 1/x. 5. cosec q = 1/y. 6. cot q = x/y. The six functions of q defined by the above equation are..

i) The numerical values of sin q and cos q cannot be greater than 1. ii) The numerical values of sec q and cosec q can never be less than 1. iii) There is no restriction on the values of tan q and cot q since they can take any valu..

In the principle value branches, the following formulae holds: - sin -1 (sin x) = x - cos -1 (cos x) = x - tan -1 (tan x) = x - cos -1 (cot x) = x - sec -1 (secx) = x - cosec -1 (cosecx) = x - sin -1 (-x) = -sin -1 x - cos -1 (-x) = p - cos -1 ..

Values of Trigonometric Functions of 30 o , 45 o , 60 o and 90 o - Let OA be the revolving ray starting from A. Let OA take the new position OP so that Draw PM perpendicular to OX and produce it to Q. Draw O..