What is 'cotx'? cot is a short way to write 'cotangent'. This is the reciprocal of the trigonometric ...

How do you simplify #(cotx + cscx) (cotx - cscx)#? Trigonometry Trigonometric Identities and Equations Fundamental Identities. 1 Answer

2018年1月22日 · #"differentiate using the "color(blue)"quotient rule"# #"given "f(x)=(g(x))/(h(x))" then"# #f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#

2015年11月18日 · Develop the expression in terms of sin x and cos x. #(cos x/sin x)(cos^2 x) + (cos x/sin x)(sin^2 x)# = #= (cos x/sin x)(cos^2 x + sin^2 x) = cos x/sin x = cot x#

2016年8月1日 · Expand and simplify LHS - Apply identity #sin^2x+cos^2x = 1#. Explanation: To verify: #(cotx - cscx)(cosx +1) =-sinx#

2015年11月3日 · Verify trig expression. (cot x/sec^2 x) + (cot x/csc^2 x) = cot x First term of left side: ((cos x/(sin x))/(1/(cos^2 x))) = (cos^3 x)/(sin x) (1) Second term of left ...

2016年7月21日 · This is an interesting problem. There are different strategies of approaching this question, but I will take what I see simplest: use of the chain rule. Let y = 3^u and u = cotx. We …

In order to derive #cot(x)/(1-sin(x))#, you need to apply the quotient rule first.. The quotient rule states this:

2016年9月26日 · dy/dx = sec^(2)x+(-csc^(2)x) The derivative is distributive over addition and subtraction. Hence: dy/dx = D_x[tanx] + D_x[cotx] Differentiating we get: dy/dx = sec^(2)x+( …

2016年7月1日 · C is the correct answer. You need to put on a common denominator. This will be sinxcosx. =(sin^2x + cos^2x)/(sinxcosx) Applying the identity sin^2x + cos^2x = 1: …