Question

(tan A/1- secA) + (1 + sec A/tan A)​

Answer 1

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(tan A / 1 - sec A)  +  (1 + sec A / tan A)

[Tan A = Sin A/Cos A    and     Sec A = 1/Cos A]

$(\frac{Sin A}{Cos A}/ 1 - \frac{1}{Cos A} ) + ( 1+\frac{1}{Cos A} /\frac{Sin A}{Cos A}$)

$=(\frac{Sin A}{Cos A} /\frac{Cos A - 1}{Cos A})+ (\frac{CosA+1}{Cos A} /\frac{Sin A}{Cos A})\\\\\\=(\frac{Sin A }{Cos A} * \frac{Cos A}{Cos A-1} )+(\frac{CosA+1}{CosA} *\frac{CosA}{SinA})\\\\\\ =\frac{Sin A}{CosA-1} +\frac{CosA+1}{SinA}$

$=\frac{Sin^2A+Cos^2A-1}{Sin A(CosA-1)}$

Sin^2 A+Cos^2 A = 1

$=\frac{1-1}{SinA(CosA-1)} \\\\\\= \frac{0}{SinA(CosA-1)} \\\\\\=0$

Therefore, the answer is 0