cosecant

put cosecx + cotx = t =- log t = - log |cosec x +cot x | \ The solution is y= log |cosec x+ cot x| + 2x+ c 2 Giving at x = p /2 = 0, y =..

1. 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 2. If a is some constant angle, then - sin q = sin a q = n p +..

Evaluate ∫ cos 3 x + cos 5 x sin 2 x + sin 4 x d x . => sin x - 2 cosec x - 6 tan -1 (sin x ) + C or sin x - 6 tan -1 (sin x ) + C or sin x - 2 (sin x ) -1 -5 tan -1 (sin x ) + C or sin x - 2(sin x ) -1 + C..

Evaluate: ∫ (9 cot m - 7 sin 6 m cos m ) dm => 9 ln | sin m | - sin 7 m + C or 9 ln | sin m | + sin 7 m + C or 9 ln | sin m | - 6 sin 7 m + C or - 9 cosec 7 m - sin 7 m + C or 9 ln | sin m | - m 7 + C..

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..

(iii) Tangent q : It is defined as the ratio of the side opposite to q and the adjacent side, In short, Similarly three more ratios can be obtained by taking the reciprocals of . (iv) Cosecant q is the reciprocal of sin It is written as cosec q..

(iii) tangent A tangent A In short, tan A Similarly three more ratios can be obtained by taking the reciprocals of sine, cosine and tangent ratios. (iv) cosecant In short, (v) seca..

(iii) tangent A tangent A In short, tan A Similarly three more ratios can be obtained by taking the reciprocals of sine, cosine and tangent ratios. (iv) cosecant In shor..

Here cosec 0 o and cot 0 o are not define..