Question

1 + cos theta + sin theta upon 1 + cos theta minus sin theta equal to 1 + sin theta upon cos theta.
(instead of dividing the nr and Dr by cos theeta I need any other alternative answer)

Answer 1

see this is your answer check it out

Answer 2

Step-by-step explanation:

Note: For the better understanding, I am replacing θ with A.

$Given:\frac{1+cosA+sinA}{1+cosA-sinA}$

On rationalizing, we get

$=\frac{1+cosA+sinA}{1+cosA-sinA}*\frac{1+cosA+sinA}{1+cosA+sinA}$

$=\frac{[(1 + cosA) + sinA]^2}{(1 + cosA)^2-(sin^2A)}$

∴ (a + b)² = a² + b² + 2ab and (a + b)(a - b) = a² - b²

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

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

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

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

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

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

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

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

Hope it helps!