Question

Pls..prove this....[tanx secx -1] / [tanx -secx 1]= tanx secx ....

i am confused

Answer 1

$\huge{Hey Mate!!!}$

Here is ur answer ⏬⏬⏬⏬⏬

❇➡  (tan x+sec x-1)/(tan x-sec x+1)

=(sin x/cos x+1/cos x-1)/(sin x/cos x-1/cos x+1)          [tan x=sin x/cos x &sec x=1/cos x]

={(sin x+1-cos x)/cos x}/{(sin x-1+cos x)/cos x}

=(sin x-cos x+1)/(sin x+cos x-1)

={(sin x-cos x+1)+(sin x+cos x-1)}/{(sin x-cos x+1)-(sin x+cos x-1)}    [by componendo dividendo]

=(sin x-cos x+1+sin x+cos x-1)/(sin x-cos x+1-sin x-cos x+1)

=2sin x/(2-2cos x)

=2sin x/2(1-cos x)

=sin x/(1-cos x)

=2sin (x/2)cos( x/2)/[1-{1-2sin2(x/2)}]     [since,sin x=2sin (x/2)cos (x/2) ]

=2sin (x/2)cos (x/2) / 2sin2(x/2)

=cos (x/2) / sin x/2

={cos (x/2)+sin (x/2)}/{cos (x/2)-sin (x/2)}     [by componendo dividendo]

=[{cos (x/2)+sin (x/2)}{cos (x/2)+sin (x/2)}] / [{cos( x/2)-sin (x/2)}{cos (x/2)+sin (x/2)}]   [rationalise]

={cos2(x/2)+sin2(x/2)+2sin (x/2)cos (x/2)} / {cos2(x/2)-sin2(x/2)}

=(1+sin x)/cos x

=1/cos x+sin x/cos x

=sec x+tan x

=tan x+sec x

Hope it helps.

Cheers!!!!

Answer 2

$\huge{Hello Friend}$

The answer of u r question is..✌️✌️

Ans:✍️✍️✍️✍️✍️❣️✍️


$\frac{(tan \: x + secx - 1)}{(tan \: x - sec \: x + 1)}$


$= \frac{(sin \: \frac{x}{cos \: + \frac{x + 1}{cosx - 1} } )}{(sin \: \frac{x}{cos \: \frac{x - 1}{cosx + 1} }) }$


$= \frac{tan \: x = sin \: x}{cosx = sec \frac{x = 1}{cos x } }$


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

$= \frac{sin \: x - cos \: x + 1sin \: x \: + cos \: x - 1}{sin \: x \: - cos \: x + 1sinx \: - cosx + 1}$



$= \frac{2sinx}{2 - 2cos \: x}$


$= \frac{2sin \: x}{2(1 - cosx)}$


$= \frac{sinx}{1 - cosx}$


$= \frac{2sin \: \frac{x}{2} cos \: \frac{x}{2} }{1 - (1 - 2sin \: 2 \frac{x}{2}) }$


$\frac{2 \sin( \frac{x}{2} \cos( \frac{x}{2} ) }{2 \sin(2 \frac{x}{2} ) }$


$= \cos( \frac{x}{2} ) \sin( \frac{x}{2} )$


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


$= \frac{1}{ \cos(x) } + \frac{ \sin(x) }{ \cos(2) }$


$= 2\sec(x) + \tan(2)$


$= \tan(x) + \sec(x)$


$\huge{Thank you}$