squareroot of 1+siny/1-siny
the answer is: 1+siny/cosy
how did they get the answer? thanks
the answer is: 1+siny/cosy
how did they get the answer? thanks
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√(1 + sin(y))/(1 - sin(y)) = √(1 + sin(y))²/√(1 - sin²(y)) = (1 + sin(y))/cos(y)
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Squareroot 1+siny/1-siny = squareroot 1+siny / squareroot 1-siny
Multiply numerator and denominator by squareroot 1+siny = 1+siny / squareroot 1-sin^2y
1-sin^2y = cos^2y so... 1+siny / squareroot cos^2y
Squareroot of cos^2y = cosy so... 1+siny / cosy
Multiply numerator and denominator by squareroot 1+siny = 1+siny / squareroot 1-sin^2y
1-sin^2y = cos^2y so... 1+siny / squareroot cos^2y
Squareroot of cos^2y = cosy so... 1+siny / cosy
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Interpreting question as
√( (1 + sin(y)) / (1 - sin(y)) )
= |tan(y/2+π/4)|
√( (1 + sin(y)) / (1 - sin(y)) )
= |tan(y/2+π/4)|
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is the whole thing under the square root? Is your instruction to simplify? Are you proving an identity?