Question

1/sec A+ tan A =1-sin A /cos A

Answer 1

$\small \bf \blue{hey \: mate \: here \: is \: ur \: ans} \\ \\ \small \bf \blue{ \huge \: solution} \\ \\ \small \bf \blue{as \: given} \\ \\ \small \bf \blue{ \frac{1}{sec \: a} +tan \: a = \frac{1 - sin \: a}{cos \: a} } \\ \\ \small \bf \blue{taking \: rhs} \\ \\ \small \bf \blue{ \frac{1 - sin \: a}{cos \: a} } \\ \\ \small \bf \blue{ \frac{(1 - sin \: a)(1 + sin \: a)}{cos \: a(1 + sin \: a)} } \\ \\ \small \bf \blue{using \: this \: identity} \\ \\ \small \bf \blue{ {a}^{2} - {b}^{2} = (a + b)(a - b)} \\ \\ \small \bf \blue{ \frac{ {1}^{2} - sin^{2} a}{cos \: a(1 + sin \: a)} } \\ \\ \small \bf \blue{ \frac{1 - sin ^{2}a }{cos \: a(1 + sin \: a)} } \\ \\ \small \bf \blue{ \frac{cos^{2} a}{cos \: a(1 + sin \: a)} \: (hence \: sin^{2} + cos^{2} = 1)} \\ \\ \small \bf \blue{( \therefore \: cos^{2} = 1 - sin^{2}) } \\ \\ \small \bf \blue{ \frac{cos \: a \: \times cos \: a}{cos \: a(1 + sin \: a)} } \\ \\ \small \bf \blue{ \frac{cos \: a}{1 + sin \: a} } \\ \\ \small \bf \blue{ \frac{cos \: a}{1} + \frac{cos \: a}{sin \: a} } \\ \\ \small \bf \blue{ \frac{1}{sec \: a} + cot \: a \:( \: ans) }$