Answers

2015-08-07T17:03:33+05:30
Write 01 + sin(theta) as 1 + cos(90-theta) 
Now 1 + cos(90-theta) = 2*(cos(45-theta/2))^2 
( Since 1 + cos(2*theta) = 2 (cos(theta))^2 )
Similarly 1 - sin(theta) becomes equal to 2*(sin(45-theta/2))^2
Thus the overall entity under the root becomes (cot(45-theta/2))^2
Now under root (cot(45-theta/2))^2 = |cot(45-theta/2)| ( Since square root of x^2 = |x| where |.| represents modulus of "." )
Next, |cot(45-theta/2)| = |tan(45+theta/2)| ( Using cot(x) = tan(90-x) )
Assuming tan(45+theta/2) to be positive, you get the desired proof .
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