SOLUTION: If 2sec2a = tanb+cotb then smallest positive value of a+b measured in degree is equal to

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Question 906268: If 2sec2a = tanb+cotb then smallest positive value of a+b measured in degree is equal to
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
2sec(2a) = tan(b) + cot(b)

2sec%282a%29=sin%28b%29%2Fcos%28b%29%2Bcos%28b%29%2Fsin%28b%29

2sec%282a%29=%28sin%5E2%28b%29%2Bcos%5E2%28b%29%29%2F%28cos%28b%29sin%28b%29%29

2sec%282a%29=1%2F%28cos%28b%29sin%28b%29%29

Divide both sides by 2

sec%282a%29=1%2F%282cos%28b%29sin%28b%29%29

sec%282a%29=1%2F%28sin%282b%29%29

1%2Fcos%282a%29=1%2F%28sin%282b%29%29

cos%282a%29=sin%282b%29

Since we are looking for small positive values
of a and b, and since the sine of one equals 
the cosine of the other, we can take 2a and 2b 
as the acute angles of a right triangle. 

So 

2a+2b = 90°

Dividing thru by 2:

a+b = 45°

Edwin