(Cot a+ Csc B)/(tan a+ sinB) = cot a Csc B

Question

• use the receprocal identities to write the trignkmetric functions in the denominator of the left side.
• add the fraction in the denominator of the previous step...(left side)
• the fraction from the previous step the simplifies fo cotA CscB using wha identity? 

Answer 1

The three bullets above need to be written as trigonometric identities.

1) We express all functions as ratios of sinA, cosA, sinB and cosB:

 

(Cot a+ Csc B)/(tan a+ sinB)

 

= (cosA/sinA + 1/sinB)/(sinA/cosA + sinB)

 

2) we use common denominators in the numerator and the denominator to simplify.

      Numerator:

      cosA/sinA + 1/sinB

      =cosA•sinB/(sinA•sinB) + sinA/(sinA•sinB)

      =(cosA•sinB + sinA)/(sinA•sinB).

 

      Denominator:

      sinA/cosA + sinB

      =sinA/cosA +sinB•cosA/cosA

      =(sinA +cosA•sinB)/cosA.

 

3. Now, we simplify by dividing the Numerator and the Denominator from previous step. We use the fact that dividing by fraction is actually multiplying by the reciprocal of the fraction. Computations:

       Numerator/Denominator

       =[(cosA•sinB + sinA)/(sinA•sinB)] /

         [(sinA +cosA•sinB)/cosA]

 

        =[(cosA•sinB + sinA)/(sinA•sinB)] •

          [cosA/(sinA +cosA•sinB)]

 

         =[(cosA•sinB + sinA) • cosA] /

           [(sinA•sinB•(cosA•sinB +sinA)]

                                                                               We rearrange now to get

         =[cosA/sinA] • [1/sinB] •[(cosA•sinB + sinA)/(cosA•sinB + sinA)]

 

         = [cotA]       • [cscB]     •     [1]

 

          = cotA • cscB.