Prove: 1/(1-sinx) - 1/(1+sinx) = 2sinx.sec²x | Trigonometric Identity

Question: Prove: $\dfrac{1}{1 - sinx} - \dfrac{1}{1 + sinx} = 2sinx.sec^2x$.

Solution:

Taking LHS

$= \dfrac{1}{1 - sinx} - \dfrac{1}{1 + sinx}$

$= \dfrac{(1 + sinx) - (1 - sinx)}{(1-sinx)(1+ sinx)}$

$= \dfrac{1 + sinx -1 + sinx}{1 - sin^2x}$

[Using (1 - sin²x = cos²x)]

$= \dfrac{2sinx}{cos^2 x}$

$= 2sinx × \dfrac{1}{cos^2x}$

[Using 1/cosx = secx]

$= 2sinx.sec^2x$

= RHS

#proved

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