Nov 5, 2016 · How do you prove #(1-\cos^2 x)(1+\cot^2 x) = 1#? How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#?

Apr 13, 2016 · How do you prove #(1-\cos^2 x)(1+\cot^2 x) = 1#? How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#?

Sep 1, 2016 · Remember the trigonometric identities. Generally there is a set output for certain values: sin(x)= 1/(csc(x)) Thus 1/sin(x)=csc(x)

How do you prove #(1-\cos^2 x)(1+\cot^2 x) = 1#? How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#?

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

Nov 1, 2017 · How do I find the logarithm #log_(1/4) 1/64#? How do I find the logarithm #log_(2/3)(8/27)#? See all questions in Logarithm-- Inverse of an Exponential Function

Oct 6, 2016 · #(sqrt(1+sinx) -1) / (x)= ((sqrt(1+sinx) -1)(sqrt(1+sinx) +1)) /( x(sqrt(1+sinx) -1)) = (1+sinx-1)/( x(sqrt(1+sinx) +1)) = sin x/x 1/(sqrt(1+sinx) +1)#

Aug 26, 2016 · See below. Use the following identities to simplify the left-hand side: •tanbeta = sinbeta/cosbeta •secbeta = 1/cosbeta Start the simplification process: (sinx/cosx + 1/cosx)(1 - sinx) = cosx ((sinx + 1)/(cosx))(1 - sinx) = cosx (1 - sin^2x)/cosx = cosx Now, rearrange the pythagorean identity sin^2beta + cos^2beta = 1 for cos^2beta to get cos^2beta = 1- sin^2beta. Applying this to our ...

Nov 12, 2016 · How do you prove #(1-\cos^2 x)(1+\cot^2 x) = 1#? How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#?

Jan 17, 2017 · Left Hand Side: #1/(sinx+1) + 1/(cscx +1) =1/(sinx+1) +1/(1/sinx +1)# #=1/(sinx+1) +1/((1 +sinx)/sinx)# #=1/(sinx+1) +sin x/(1 +sinx)#