Yes.
yes 1 + cot x^2 = csc x^2
Ah, secant, annoying as always. Why don't we use its definition as 1/cos x and csc as 1/sin x? We will do that Also, please write down the equation, there is at least TWO different equations you are talking about. x^n means x to the power of n 1/(sin x) ^2 is csc squared x, it's actually csc x all squared 1/(cos x) ^2 in the same manner.
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
How is it possible that the value of cosecant is less than 1 (2/7)?
It is cotangent(theta).
It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)
cot(x)=1/tan(x)=1/(sin(x)/cos(x))=cos(x)/sin(x) csc(x)=1/sin(x) sec(x)=1/cos(x) Therefore, (csc(x))2/cot(x)=(1/(sin(x))2)/cot(x)=(1/(sin(x))2)/(cos(x)/sin(x))=(1/(sin(x))2)(sin(x)/cos(x))=(1/sin(x))*(1/cos(x))=csc(x)*sec(x)
negative cotangent -- dcot(x)/dx=-csc^2(x)
By converting cosecants and secants to the equivalent sine and cosine functions. For example, csc theta is the same as 1 / sin thetha.
1/sin x = csc x
The statement "cot multiplied by cosec equals cos" is not accurate. In trigonometric terms, cotangent (cot) is the reciprocal of tangent, and cosecant (cosec) is the reciprocal of sine. Therefore, the correct relationship is ( \cot(x) \cdot \csc(x) = \frac{\cos(x)}{\sin^2(x)} ), which does not simplify to cosine. Instead, it highlights the relationship between these functions in terms of sine and cosine.
pi divided by 6 is a constant and so its first derivative is 0. And since that is also a constant, the second derivative is 0. It is not clear what f(x) = csc(x) has to do with that!