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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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