Question

COSEC A -1/COSEC A +1= (COS A / 1+SIN A)2
irrevelent answer will be reprted 100%

Answer 1

Answer:$\frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}\\=2secA$

Step-by-step explanation:

$LHS = \frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}\\=\frac{cos^{2}A+(1+sinA)^{2}}{(1+sinA)cosA}\\=\frac{cos^{2}A+1^{2}+sin^{2}A+2sinA}{(1+sinA)cosA}\\=\frac{(cos^{2}A+sin^{2}A)+1+2sinA}{(1+sinA)cosA}\\=\frac{1+1+2sinA}{(1+sinA)cosA}$

/* By Trigonometric identity:

cos² A+ sin² A = 1 */

$=\frac{2+2sinA}{(1+sinA)cosA}\\=\frac{2(1+sinA)}{(1+sinA)cosA}$

After cancellation,we get

$= \frac{2}{cosA}\\=2secA\\=RHS$

Therefore,

$\frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}\\=2secA$

Answer 2

Answer:

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