Question

` 3.Prove that:- `(sin A)/(1+cos A)+(sin A)/(1-Cos A)=2cosec A​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Prove \: that : \: \dfrac{sinA}{1 + cosA} + \dfrac{sinA}{1 - cosA} = 2cosecA$

$\large\underline{\sf{Solution-}}$

Identities Used :-

$\boxed{ \bf \: {x}^{2} - {y}^{2} = (x + y)(x - y)}$

$\boxed{ \bf \: 1 - {cos}^{2} x = {sin}^{2} x}$

$\boxed{ \bf \: \dfrac{1}{sinx} = cosecx}$

Let's solve the problem now!!

Consider,

$\rm :\longmapsto\:\: \dfrac{sinA}{1 + cosA} + \dfrac{sinA}{1 - cosA}$

$\: \: \sf \: = \: \: sinA\bigg(\dfrac{1}{1 + cosA} + \dfrac{1}{1 - cosA} \bigg)$

$\: \: \sf \: = \: \: sinA\bigg(\dfrac{1 - \cancel{cosA} + 1 + \cancel{cosA}}{(1 + cosA)(1 - cosA)} \bigg)$

$\: \: \sf \: = \: \: sinA \: \bigg(\dfrac{2}{1 - {cos}^{2} A} \bigg)$

$\: \: \sf \: = \: \: sinA \times \dfrac{2}{ {sin}^{2} A}$

$\: \: \sf \: = \: \: \dfrac{2}{sinA}$

$\: \: \sf \: = \: \: 2cosec \: A$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{ \bf \: {sin}^{2} x + {cos}^{2} x = 1}$

$\boxed{ \bf \: {sec}^{2} x - {tan}^{2} x = 1}$

$\boxed{ \bf \: {cosec}^{2} x - {cot}^{2} x = 1}$

$\boxed{ \bf \: secx = \dfrac{1}{cosx}}$

$\boxed{ \bf \: tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }$

$\boxed{ \bf \: cotx = \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }$