Question

√1-sinx/1+sinx=secx - tanx

Answer 1

I hope this will satisfy you

Answer 2

Answer:

$\sqrt{\frac{1-sinx}{1+sinx}}=secx-tanx$

Step-by-step explanation:

$\sqrt{\frac{1-sinx}{1+sinx}}$

$=\sqrt{\frac{(1-sinx)(1-sinx)}{(1+sinx)(1-sinx)}}$

$=\sqrt\left({\frac{(1-sinx)^{2}}{(1^{2}-sin^{2}x)}}\right)}$

/* By algebraic identity:

(a+b)(a-b)=a²-b² */

$=\sqrt{\left(\frac{(1-sinx)^{2}}{cos^{2}x}}\right)}$

$=\frac{1-sinx}{cosx}$

$=\frac{1}{cosx}-\frac{sinx}{cosx}$

$=secx-tanx\\=RHS$

Therefore,

$\sqrt{\frac{1-sinx}{1+sinx}}=secx-tanx$

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