Science Lair: Complex Trig Identities Proved

Solving Complex Trig I dentit ies.

Hi viewers, this is my third post on Trig Identities, I am fed up with this topic, anyways lets start, many still complain that they still face challenges with trig identities, so I pick some difficult(complex) questions for us to solve. Remember if you haven't read my previous posts try to check them out.
Now lets begin

1. Prove 1+cosx/1-cosx ➕ 1-cosx/1+cosx = 4cot²x +2
Solution.
When u see a fraction you try your best to get a common denominator which in this case we need to find the lowest common multiple (LCM). When there is nothing common in your denominators;the lcm is the product of the denominator.

(1+cosx)(1+cosx) ➕ (1-cosx)(1-cosx) /(1+cosx)(1-cosx).
Expanding we have
2+2cosx+2cos²x - 2cosx/1-cos²x.
2cosx and -2cosx with subtract themselves out, we have :
2+2cos²x/1-cos²x
but 1-cos²x=sin²x
if are confuse how read my post on Introduction to Trig identities.
2+2cos²x/sin²x
2(1+cos²x)/sin²x
2[1/sin²x +cos²x/sin²x]
2(cosec²x+cot²x)
but cosec²x=1 +cot²x
2(1+cot²x+cot²x)
2(1+2cot²x)=2+4cot²x=4cot²x +2.

2. Prove cotA+tanB/cotB+tanA = cotAtanB.
Now remember cotA=cosA/sinA, tanA=sinA/cosA, tanB=sinB/cosB, cotB=cosB/sinB.
substituting we have.
cosAsinB/sinAcosB
cotAtanB.

Finding the lcm we have that, now we want to simply the expression, the big division sign will turn to small division and the fraction at the Right hand side will turn upside down as the division sign will turn to multiplication