Question

Prove that tanA + tanB/cotA + cot B = tan A tan B​

Answer 1

Answer:

Since cotA = 1/tanA, substituting to get everything in terms of tanA and tanB, then (tanA + tanB)(1 - cotAcotB) = (tanA + tanB)(1 - 1/(tanAtanB) =

((tanA + tanB)(tanAtanB - 1))/(tanAtanB) - (i)

(cotA + cotB)(1 - tanAtanB) = (1/tanA + 1/tanB)(1 - tanAtanB) =

((tanA + tanB)(1 - tanAtanB))/(tanAtanB) - (ii)

Adding (i) + (ii) you get 0, since (i) = - (ii)

Answer 2

Answer:

First, chose one of the variables to isolate.

In this case, let’s isolate x . In order to do so, we need to cancel out the y variable. Multiply the first equation by 2 and the second by 3 to get the following:

4x−6y=22

15x+6y=54

Now adding both equations would rid us of the y variable to get:

19x=76 hence x=4

Now replace x by 4 in any of the original equations:

5(4)+2y=18

20+2y=18

2y=−2

y=−1

So the answer to the equations is 4;−1