Prove that `(1+sinx)/(cosx) + (cosx)/(1-sinx) = 2sec x`
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`(1+sinx)/(cos x)+ (cosx)/(1-sinx) = 2 sec x`
Let's simplify left side of the equation.
Since the denominators are cos x and 1-sin x, the LCD is cosx(1-sinx).
`(1+sinx)/(cosx) *(1-sinx)/(1-sinx) + (cosx)/(1-sinx)*(cosx)/(cosx)`
`=(1-sin^2x) /(cos x(1-sinx)) + (cos^2x)/(cosx(1-sinx))`
Base on the Pythagorean identity, `cos^2x = 1- sin^2x` .
`= (1-sin^2x)/(cosx(1-sinx)) + (1-sin^2x)/(cos x(1-sinx))`
Add the fractions.
`= (2(1-sin^2x))/(cosx(1-sinx) )`
Factor `1-sin^2x` .
`= (2(1-sinx)(1+sinx))/(cosx(1- sinx))`
Cancel common factor between numerator and denominator.
`= (2 (1+sinx))/(cosx)`
`= (2+2sinx)/(cosx)`
Express as two fractions.
`= 2/(cosx) + (2sinx)/(cosx)`
Note that `sec x = 1/(cos x)` and `tanx =(sinx)/(cosx)` .
`= 2secx + 2tanx`
Since the simplified form of the left side is not the same with the given expression at the right side of the equation, hence
`(1+sinx)/(cosx) + (cosx)/(1-sinx)` `!=` `2secx` .
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