Question

If tan =20/21 show that. (1-sin+cos)/(1+sin+cos)=3/7

Qno 11.

Answer 1

${\underline{\sf{Given}}}$

If tanx = 20/21

${\underline{\sf{To\:Prove}}}$

$\sf \dfrac{1 - \sin \: x + \cos \: x}{1 + \sin \: x + \cos \: x } = \dfrac{3}{7}$

${\underline{\sf{Solution}}}$

We have tanx = 20/21..(1)

$\sf \implies \sec \: x = \sqrt{1 + \tan {}^{2}x } = \sqrt{1 + (\frac{20}{21} ){}^{2} }$

$\sf = \sqrt{ \dfrac{841}{441} } = \dfrac{29}{21} ...(2)$

$\rule{200}2$

$\sf \: LHS = \dfrac{1 - \sin \: x + \cos \: x}{1 + \sin \: x + \cos \: x }$

Now divide numerator and denominator by cos x

$\sf = \dfrac{ \frac{1}{ \cos \:x} - \frac{ \sin \: x}{ \cos \: x} + \frac{ \cos \: x}{ \cos \: x} }{ \frac{1}{ \cos \:x} + \frac{ \sin \: x}{ \cos \: x} + \frac{ \cos \: x}{ \cos \: x}}$

$\sf = \dfrac{ \sec \: x - \tan \: x + 1}{ \sec \: x + \tan \: x + 1}$

Now put the values of secx and tan x

$\sf = \dfrac{ \frac{29}{21} - \frac{20}{21} + 1}{\frac{29}{21} + \frac{20}{21} + 1}$

$\sf = \dfrac{30}{70}$

$\sf = \dfrac{3}{7}$

$\sf = RHS$

Hence proved !