Solveeit Logo
Login

Question

Question: .\(\frac{\cos A}{1 - \sin A} =\)...

.cos⁡A1−sin⁡A=\frac{\cos A}{1 - \sin A} =

A

sec⁡A−tan⁡A\sec A - \tan A

B

cosecA+cot⁡A\text{cosec}A + \cot A

C

tan⁡(π4−A2)\tan\left( \frac{\pi}{4} - \frac{A}{2} \right)

D

tan⁡(π4+A2)\tan\left( \frac{\pi}{4} + \frac{A}{2} \right)

Answer

tan⁡(π4+A2)\tan\left( \frac{\pi}{4} + \frac{A}{2} \right)

Explanation

Solution

cos⁡A1−sin⁡A=cos⁡A(1+sin⁡A)cos⁡2A=(1+sin⁡A)cos⁡A\frac{\cos A}{1 - \sin A} = \frac{\cos A(1 + \sin A)}{\cos^{2}A} = \frac{(1 + \sin A)}{\cos A}

=(cos⁡A2+sin⁡A2)2(cos⁡A2+sin⁡A2)(cos⁡A2−sin⁡A2)=cos⁡A2+sin⁡A2cos⁡A2−sin⁡A2= \frac{\left( \cos\frac{A}{2} + \sin\frac{A}{2} \right)^{2}}{\left( \cos\frac{A}{2} + \sin\frac{A}{2} \right)\left( \cos\frac{A}{2} - \sin\frac{A}{2} \right)} = \frac{\cos\frac{A}{2} + \sin\frac{A}{2}}{\cos\frac{A}{2} - \sin\frac{A}{2}}

=1+tan⁡A21−tan⁡A2= \frac{1 + \tan\frac{A}{2}}{1 - \tan\frac{A}{2}}, (DividingNrandDrbycos⁡A2)\left( \text{Dividing}N^{r}\text{and}D^{r}\text{by}\cos\frac{A}{2} \right)

=tan⁡(π4+A2)= \tan\left( \frac{\pi}{4} + \frac{A}{2} \right).