Question

1+cosx/sinx + sinx/ 1+cosx = 2/sinx

Plz answer it's urgent!!

The one who answers first I will mark as brainliest!!​

Answer 1

TO PROVE :–

$\\ \implies \sf \dfrac{1 +\cos(x) }{ \sin(x) } +\dfrac{ \sin( x) }{1 +\cos(x) } = \dfrac{2}{ \sin(x) }$

SOLUTION :–

• Let's take L.H.S. –

$\\ = \sf \dfrac{1 + \cos(x) }{ \sin(x) } + \dfrac{ \sin( x) }{1 + \cos(x)}$

$\\ = \sf \dfrac{ [1 + \cos(x)]^{2}+ \sin^{2} (x)}{ \sin(x) [1 + \cos(x)]}$

• Using identity –

$\\ \longrightarrow \sf {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

$\\ = \sf \dfrac{ 1 + \cos ^{2} (x)+ 2 \cos(x) + \sin^{2} (x)}{ \sin(x) [1 + \cos(x) ] }$

$\\ = \sf \dfrac{ 1 + 2 \cos(x)+[ \cos ^{2} (x) + \sin^{2} (x)] }{\sin(x) [1 + \cos(x) ] }$

$\\ = \sf \dfrac{ 1 + 2 \cos(x)+[ \cos ^{2} (x) + \sin^{2} (x)] }{ \sin(x) [1 + \cos(x) ] }$

$\\ = \sf \dfrac{1 + 2 \cos(x)+1 }{ \sin(x) [1 +\cos(x) ] } \:\:\:\:\:\: \: \: \: \: \:\: [ \: \because \:\:\cos ^{2} (x)+ { \sin}^{2} (x) = 1]$

$\\ = \sf \dfrac{ 2+ 2 \cos(x) }{ \sin(x) [1 + \cos(x) ] }$

$\\ = \sf \dfrac{ 2 \cancel{[1 + \cos(x) ]}}{ \sin(x) \cancel{[1 + \cos(x) ] }}$

$\\ =\sf \dfrac{ 2}{ \sin(x)}$

$\\ \sf \: = R.H.S . \:\:\:\: (Hence \:\: proved)$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$: \implies{\sf{ \frac{1 + cosx}{snx} + \frac{sinx}{1 + cosx} = \frac{2}{sinx} }}$

${\bf{\blue{\underline{Now:}}}}$

Take L.HS

$: \implies{\sf{ L.H.S = \frac{1 + cosx}{sinx} + \frac{sinx}{1 + cosx} }}$

$= {\sf{ \frac{(1 + cosx)(1 + cosx) + (sinx)(sinx)}{(sinx)(1 + cosx)} }}$

$= {\sf{ \frac{(1 + cosx)^{2} + (sinx)^{2} }{(sinx)(1 + cosx)} }}$

$\dagger \: \: \boxed{\sf{ ( {x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy }}$

$= {\sf{ \frac{ {(1)}^{2} + {cos}^{2}x + 2cosx + {sin}^{2} x }{sinx(1 + cosx)} }}$

$= {\sf{ \frac{ 1 + ({cos}^{2}x + {sin}^{2} x ) + 2cosx}{sinx(1 + cosx)} }}$

$\dagger \: \: \boxed{\sf{ \purple{ {sin}^{2} x + {cos}^{2} x= 1 }} }$

$= {\sf{ \frac{1 + 1 + 2cosx}{sinx(1 + cosx)} }}$

$= {\sf{ \frac{2 + 2cosx}{sinx(1 + cosx)} }}$

Take 2 common,

$= {\sf{ \frac{2(1 + cosx)}{sinx(1 + cosx)} }}$

$= {\sf{ \frac{2 \cancel{(1 + cosx)}}{sinx \cancel{(1 + cosx)}} }}$

$= {\sf{ \frac{2}{sinx} }}$

Hence L.HS=R.HS