How do you simplify the expression $\frac{1}{sec t - tan t} - \frac{1}{sec t + tan t}$ $1/(sect-tant)-1/(sect+tant)$ ?

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Answer 1 · 2016-09-26T10:12:02

$\frac{1}{sec t - tan t} - \frac{1}{sec t + tan t} = 2 tan t$ $1/(sect-tant)-1/(sect+tant)=2tant$

$\frac{1}{sec t - tan t} - \frac{1}{sec t + tan t}$ $1/(sect-tant)-1/(sect+tant)$

= $\frac{(sec t + tan t) - (sec t - tan t)}{(sec t - tan t) (sec t + tan t)}$ $((sect+tant)-(sect-tant))/((sect-tant)(sect+tant))$

= $\frac{sec t + tan t - sec t + tan t}{{sec}^{2} t - {tan}^{2} t}$ $(sect+tant-sect+tant)/(sec^2t-tan^2t)$

= $\frac{2 tan t}{1} = 2 tan t$ $(2tant)/1=2tant$

How do you simplify the expression 1/(sect-tant)-1/(sect+tant)1sect−tant−1sect+tant1/(sect-tant)-1/(sect+tant)?