Question

hey guys ...
its urgent tommarow is my exam​

Answer 1

Answer:

Step-by-step explanation:

We have $\sqrt \frac{1-\sin{A}}{1+\sin{A}} = \sec{A} - \tan{A}$

Consider the LHS

$\sqrt \frac{1-\sin{A}}{1+\sin{A}}$

Multiply and Divide by $\sqrt{(\,1-\sin{A})\,}$ , we get

$\sqrt \frac{1-\sin{A}}{1+\sin{A}} \times \sqrt{\frac{1 - \sin{A}}{1 - \sin{A}}}\\\\ \sqrt \frac{(\,1 + \sin{A})\,^2}{(\,1)\,^2 - (\,\sin{A})\,^2}\\\\ \sqrt{\frac{(\,1 + \sin{A})\,^2}{1 - \sin{A}^2}}\\\\ \sqrt{\frac{(\,1 + \sin{A})\,^2}{\cos{A}^2}}\\\\ \frac{1 + \sin{A}}{\cos{A}}\\\\ \frac{1}{\cos{A}} + \frac{\sin{A}}{\cos{A}}\\\\ \sec{A} + \tan{A}$

Thus, we have LHS = RHS

Answer 2

hey here is your answer...

(¡) &(¡¡)answer....

hope this helps you

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