Question

find the (tan+sec-1/tan-sec+1)1/sec​

Answer 1

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

Hopes it Helps

Answer 2

Solution :

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

$\implies( \frac{(tan +sec) - ( {sec}^{2} - {tan}^{2}) }{tan \: - sec \: + 1} ) \frac{1}{sec}$

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

$\implies \: (tan + sec) \frac{1}{sec}$

$\implies \: ( \frac{sin \: + 1}{cos} ) \frac{1}{sec}$

$\implies \: \frac{1 + sin}{cos \: \times sec}$

$\implies1 + sin$