2024年6月11日 · cosec x = 1/sinx = sinx/sin²x = (sin x)/ (1 - cos²x). Let u = cos x and then proceed by partial fractions after the substitution . Hope this helps! ~~Simba

2018年4月26日 · I got 2 (sec^2 x + cosec^2 x) instead of it. Is the question wrong or something else? It is an IGCSE past paper question (Additional Math)

2017年1月13日 · Given that cos (x) = -3/5 and x lies in the 2nd quadrant, what is the value of sec (x), cosec (x) and cot (x)?

2018年8月13日 · The complete solution set of the equation 1+2cosecx= (sec^2x/2)/2 is what?

2024年6月11日 · Hi, I was revising for an upcoming C3 paper when i came across this: Prove that: (given the below part (a) to the question) cosec^4 (x) cosec4(x)- cot^4 (x) cot4(x)= cosec^2 (x) cosec2(x) + cot^2 (x) cot2(x) well bascially I wasnt sure how to tackle this one so I did a bit or rearranging to both sides to give:

2018年4月4日 · We seek to prove the identity: cosec(x −B) sec(x + B) = 1 − tanxtanB tanx − tanB Consider the LHS: LH S = cosec(x −B) sec(x + B) Using the fundamental definitions of trigonometric functions:

2024年5月22日 · Hi i no from formula sheets that the integral of cosec^2x equals -cotx but i still cant work out how to prove this integral. so far iv tried a few different things 1. i no that cosec^2=1/sin^2x but this doesnt help much 2. cosec^2x=cot^2x + 1, again... couldn't get much further with that any help would be greatly appreciated thanks

Do you know the quotient rule? Because you use it to find the derivative of tan x and cot x by writing them in terms of sin x and cos x. Also, you can use the chain rule to find the derivatives of sec x and cosec x, since \sec x = (\cos x)^ {-1} secx = (cosx)−1, and so on.

2017年3月10日 · LH S = (1 + tanx sinx) + (1 + cotx cosx) = 1 sinx + sinx cosxsinx + 1 cosx + cosx sinxcosx = cscx + secx + secx + cscx = 2(secx + cscx) = RH S

2021年8月30日 · I'm getting rusty in my old age.... anyone have any ideas on how to prove the following identity? (cosec x - cot x)/ (1 - cos x) = sec x I've tried multiple methods but always seem to get 1/sin x rather than sec x.