Question #c44b0

2 Answers
Jan 11, 2018

Start by using #cotx = cosx/sinx# and #tanx = sinx/cosx#. My first step in proving identities is usually to rewrite in sine and cosine.

#1/(sinA/cosA + cosA/sinA) = sinAcosA#

#1/((sin^2A+cos^2A)/(cosAsinA)) = sinAcosA#

Now use #sin^2x +cos^2x =1#.

#cosAsinA = sinAcosA#

Hopefully this helps!

Jan 11, 2018

On verifies an equation of this type by using identities and axioms to change only one side until it is identical to the other side.

Explanation:

Verify: #1/(tan(A) + cot(A))= sin(A)cos(A)#

Substitute #tan(A) = sin(A)/cos(A)# and #cot(A) = cos(A)/sin(A)#:

#1/(sin(A)/cos(A) + cos(A)/sin(A))= sin(A)cos(A)#

Make a common denominator for the fractions of the denominator:

#1/(sin(A)/cos(A)sin(A)/sin(A) + cos(A)/sin(A)cos(A)/cos(A))= sin(A)cos(A)#

Perform the multiplication:

#1/(sin^2(A)/(sin(A)cos(A)) + cos^2(A)/(sin(A)cos(A)))= sin(A)cos(A)#

Combine the numerators:

#1/((sin^2(A) + cos^2(A))/(sin(A)cos(A)))= sin(A)cos(A)#

Substitute #sin^2(A)+cos^2(A) = 1#:

#1/(1/(sin(A)cos(A)))= sin(A)cos(A)#

Division by a fraction is the same as multiplication by the reciprocal of the divisor:

#sin(A)cos(A)= sin(A)cos(A)#