Question

prove that 1+ sinx- cosx/ 1+ sinx + cos x= tan x/2​

Answer 1

GIVEN:

• (1 + sinx - cosx) / (1 + sinx + cos x) = tan x/2

TO PROVE:

• (1 + sinx - cosx) / (1 + sinx + cos x) = tan x/2

EXPLANATION:

$\boxed{ \bold{ \large{ \gray{Sin \ 2x = 2sin \ x \ cos \ x}}}}$

$\sf Sin \ x = 2sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \ cos \ \left \lgroup\dfrac{x}{2} \right \rgroup$

$\boxed{ \bold{ \large{ \gray{Cos\ 2x = 2cos^2\ x - 1}}}}$

$\sf Cos\ x = 2cos^2\ \left \lgroup\dfrac{x}{2} \right \rgroup - 1$

L.H.S : (1 + sinx - cosx) / (1 + sinx + cos x)

$\sf \leadsto\dfrac{1 + sinx - cosx}{1 + sinx + cos x}$

Substitute the value of sin x in both numerator and denominator and the value of cos x in denominator.

$\sf \leadsto\dfrac{1 + 2sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \ cos \ \left \lgroup\dfrac{x}{2} \right \rgroup - cos \ x}{1 + 2sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \ cos \ \left \lgroup\dfrac{x}{2} \right \rgroup + 2cos^2\ \left \lgroup\dfrac{x}{2} \right \rgroup - 1}$

$\sf \leadsto\dfrac{1 + 2sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \ cos \ \left \lgroup\dfrac{x}{2} \right \rgroup - cos \ x}{2sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \ cos \ \left \lgroup\dfrac{x}{2} \right \rgroup + 2cos^2\ \left \lgroup\dfrac{x}{2} \right \rgroup }$

$\boxed{ \bold{ \large{ \gray{Cos\ 2x = 1 - 2sin^2\ x }}}}$

$\sf Cos\ x = 1 - 2sin^2\ \left \lgroup\dfrac{x}{2} \right \rgroup$

Substitute this value for cos x in numerator.

$\sf \leadsto\dfrac{1 + 2sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \ cos \ \left \lgroup\dfrac{x}{2} \right \rgroup - 1 + 2sin^2\ \left \lgroup\dfrac{x}{2} \right \rgroup }{2sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \ cos \ \left \lgroup\dfrac{x}{2} \right \rgroup + 2cos^2\ \left \lgroup\dfrac{x}{2} \right \rgroup }$

$\sf \leadsto\dfrac{2sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \ cos \ \left \lgroup\dfrac{x}{2} \right \rgroup+ 2sin^2\ \left \lgroup\dfrac{x}{2} \right \rgroup }{2sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \ cos \ \left \lgroup\dfrac{x}{2} \right \rgroup + 2cos^2\ \left \lgroup\dfrac{x}{2} \right \rgroup }$

$\sf \leadsto\dfrac{sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \ cos \ \left \lgroup\dfrac{x}{2} \right \rgroup+ sin^2\ \left \lgroup\dfrac{x}{2} \right \rgroup }{sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \ cos \ \left \lgroup\dfrac{x}{2} \right \rgroup + cos^2\ \left \lgroup\dfrac{x}{2} \right \rgroup }$

$\sf \leadsto\dfrac{sin \ \left \lgroup\dfrac{x}{2} \right \rgroup \left(cos \ \left \lgroup\dfrac{x}{2} \right \rgroup+ sin\ \left\lgroup\dfrac{x}{2} \right \rgroup \right) }{cos \ \left \lgroup\dfrac{x}{2} \right \rgroup \left(sin \ \left \lgroup\dfrac{x}{2} \right \rgroup + cos\ \left \lgroup\dfrac{x}{2} \right \rgroup \right) }$

Cancel sin (x/2) and cos (x/2) in both numerator and denominator.

$\sf \leadsto\dfrac{sin \ \left \lgroup\dfrac{x}{2} \right \rgroup }{cos \ \left \lgroup\dfrac{x}{2} \right \rgroup}$

$\boxed{ \bold{ \large{ \gray{\dfrac{Sin \ x }{Cos \ x}= Tan x}}}}$

$\sf \leadsto\dfrac{sin \ \left \lgroup\dfrac{x}{2} \right \rgroup }{cos \ \left \lgroup\dfrac{x}{2} \right \rgroup} = tan \ \left \lgroup\dfrac{x}{2} \right \rgroup$

R.H.S : Tan (x/2)

L.H.S = R.H.S

$\sf \leadsto\dfrac{1 + sinx - cosx}{1 + sinx + cos x} = tan \ \left \lgroup\dfrac{x}{2} \right \rgroup$

HENCE PROVED.