Question

tan0+sec0-1/tan0-sec0+1=1+sin0/cos0

Answer 1

$\huge{\bigstar}\huge{\mathcal{ \overbrace{ \underbrace{ \pink{ \fbox{ \green{ \blue{S} \pink{o} \red{l} \green{u} \purple{t} \blue{i} \green{o} \red{n}}}}}}}} \huge{ \bigstar}$

$\purple{ \underline{ \underline{ \mathbb{L.H.S.}}} } \\ \\ \bf{ \orange{Formulas \: used : - }}\\ \boxed{ \bf{ {sec}^{2} \theta - {tan}^{2} \theta = 1 }} \\ \boxed{ \bf{ {a}^{2} - {b}^{2} = (a + b)(a - b) }} \\ \\ \bf{ \red{(i) \: \frac{tan \theta + sec \theta - 1}{tan \theta - sec \theta + 1} }} \\ = \bf{ \frac{tan \theta + sec \theta - ( {sec}^{2} \theta - {tan}^{2} \theta )}{tan \theta - sec \theta + 1} } \\ = \bf{ \frac{(sec \theta + tan \theta) - (sec \theta + tan \theta)(sec \theta - tan \theta)}{1 - sec \theta + tan \theta} } \\ = \bf{ \frac{(sec \theta + tan \theta)(1 - sec \theta + tan \theta)}{(1 - sec \theta + tan \theta)} } \\ = \bf{sec \theta + tan \theta} \\ = \bf{ \frac{1}{cos \theta} + \frac{sin \theta}{cos \theta} } \\ = \bf{ \frac{1 + sin \theta}{cos \theta} } = \purple{ \underline{ \underline{ \mathbb{R.H.S.}}}}$