How do you prove $( 1 / (secx - tanx) ) - ( 1 / (secx + tanx ) ) = 2 tan x$ ?

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Answer 1 · 2015-07-04T05:19:18

It can be proved by $sec^2x-tan^2x=1$

$sec^2x-tan^2x=1$

$(secx+tanx)(secx-tanx)=1$ [since $a^2-b^2=(a+b)(a-b)]$

$secx+tanx=1/(secx-tanx)$ $" " color(red)((1))$

$secx-tanx=1/(secx+tanx)$ $" " color(red)((2))$

$LHS= (1/(secx−tanx)−1/(secx+tanx))$

$" " color(red)((1))$ & $color(red)((2))$ substitute in the above equation

$LHS=(secx+tanx)-(secx-tanx)$

$LHS=secx+tanx-secx+tanx$

$LHS=2tanx$

$LHS=RHS$

Hence proved

How do you prove ( 1 / (secx - tanx) ) - ( 1 / (secx + tanx ) ) = 2 tan x ( 1 / (secx - tanx) ) - ( 1 / (secx + tanx ) ) = 2 tan x?