Question

1+sin/1-sin-1-sin/1+sin = 4 tan sec​

Answer 1

Answer:

1+sinx/1-sinx - 1-sinx/1+sinx

=(1+sinx)²-(1-sinx)²/1-sin²x

4sinx/cos²x

=4tanx secx

Step-by-step explanation:

(1+sinx)²-(1-sinx)² is in a²-b²

so we write it as (a+b)(a-b) and get 2sin X and 1-sin²x=cos²x

Answer 2

Answer:

Given,

$\frac{1 + \sin(x) }{1 - \sin(x) } - \frac{1 - \sin(x) }{1 + \sin(x) }$

We are simplifying it,

$= \frac{(1 + \sin(x))(1 + \sin(x)) }{(1 - \sin(x))(1 + \sin(x) ) } - \frac{(1 - \sin(x))(1 - \sin(x)) }{(1 + \sin(x))(1 - \sin(x) ) }$

$= \frac{ {(1 + \sin(x)) }^{2} }{ {1}^{2} - {sin}^{2} x } - \frac{ {(1 - \sin(x)) }^{2} }{ {1}^{2} - {sin}^{2}x }$

$= \frac{ {(1 + \sin(x)) }^{2} }{ {cos}^{2} x } - \frac{ {(1 - \sin(x)) }^{2} }{ {cos}^{2}x }$

$= \frac{ {1}^{2} + 2 \sin(x) + {sin}^{2}x - {1}^{2} + 2 \sin(x) + {sin}^{2}x }{ {cos}^{2}x }$

$= \frac{4 \sin(x) }{ {cos}^{2} x} \\ = 4 \frac{ \sin(x) }{ \cos(x) } \times \frac{1}{ \cos(x) } \\ = 4 \tan(x) \sec(x)$

Therefore,

$\frac{1 + \sin(x) }{1 - \sin(x) } - \frac{1 - \sin(x) }{1 + \sin(x) } = 4 \tan(x) \sec(x)$

Here applied formula,

${sin}^{2} x + {cos}^{2} x = 1$

Some more important trigonometry formula,

$sin(x) = \cos(\frac{\pi}{2} - x) \\ \tan(x) = \cot(\frac{\pi}{2} - x) \\ \sec(x) = \csc(\frac{\pi}{2} - x) \\ \cos(x) = \sin(\frac{\pi}{2} - x) \\ \cot(x) = \tan(\frac{\pi}{2} - x) \\ \csc(x) = \sec(\frac{\pi}{2} - x)$

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