Question

(tan x)/(sec x - 1) + (tan x)/(sec x + 1) is
equal to​

Answer 1

Answer:

Please make like me and hope you understand

Answer 2

$\frac{ \tan(x) }{ \sec(x) - 1 } + \frac{ \tan(x)}{ \sec(x) + 1} \\ \\ \frac{ \tan \: x (( \sec x + 1) + ( \sec \: x - 1)) }{(\sec x + 1) \: (\sec x - 1)} \\ \\ \frac{ \tan \: x \: (2 \sec \: x) }{ {sec}^{2}x - 1} \\ \\ \frac{ \tan \: x \: . \: 2 \sec \: x}{ {tan}^{2}x } \\ \\ \frac{2sec \: x}{tan \: x} \\ \\ \frac{2cos \: x}{sin \: x \: . \: cos \: x} \\ \\ = \: \: 2 \:cosec(x)$

Formula used :- sec^2x - 1 = tan^2x.

Hope this helps.