Question

Please can I have an answer?

Answer 1

Solution:-

$\large \bold{ \tt{ \frac{sinA }{1 + cosA} + \frac{1 +cosA }{sinA} }}$

$= \large \color{blue}\bold{ \tt{ \frac{ sinA \times sin A \: + (1 + cosA)(1 + cosA)}{(1 + cosA) \: sinA}}}$

$= \large \color{blue}\bold{ \tt{ \frac{ {sin}^{2} A \: + {(1 + cosA)}^{2} }{(1 + cosA) \: sinA}}}$

$= \large \color{blue}\bold{ \tt{ \frac{ {sin}^{2} A \: + { {1}^{2} + 2 \times 1 \times cosA + {cosA}^{2} }}{(1 + cosA) \: sinA}}}$

$= \large \color{blue}\bold{ \tt{ \frac{ {sin}^{2} A \: + { {1}^{} + 2cosA + {cosA}^{2} }}{(1 + cosA) \: sinA}}}$

$= \large \color{blue}\bold{ \tt{ \frac{ {sin}^{2} A \: + {cosA}^{2} + 2cosA + 1 }{(1 + cosA) \: sinA}}}$

$= \large \color{blue}\bold{ \tt{ \frac{ 1 + 2cosA + 1 }{(1 + cosA) \: sinA}}}$

$= \large \color{blue}\bold{ \tt{ \frac{ 2+ 2cosA }{(1 + cosA) \: sinA}}}$

$= \large \color{blue}\bold{ \tt{ \frac{ 2(1+ cosA) }{(1 + cosA) \: sinA}}}$

$= \large \color{blue}\bold{ \tt{ \frac{ 2 }{ \: sinA}}}$

$= \large \underbrace{\color{blue}{\bold{ 2 \: cosecA}}}$

Answer:- (4) 2cosecA