Question

prove that sinA÷1+cosA=CosecA-cotA​

Answer 1

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Answer 2

$\small \bf \pink{hey \: mate \: here \: is \: ur \: ans} \\ \\ \small \bf \pink{ \huge \: solution} \\ \\ \small \bf \pink{ \frac{sin \: a}{1 + cos \: a} =cosec \: a - cot \: a } \\ \\ \\ \\ \small \bf \pink{taking \:lhs }\\ \\ \small \bf \pink{ \frac{sin \: a}{1 + cos \: a} } \\ \\ \small \bf \pink{ \frac{sin \: a(1 - cos \: a)}{(1 + cos \: a)(1 - cos \: a)} } \\ \\ \small \bf \pink{ \frac{sin \: a(1 - cos \: a)}{ {1}^{2} - {cosa}^{2} a} } \\ \\ \small \bf \pink{ \frac{sin \: a(1 - cos \: a)}{1 - {cos \: }^{2} a} } \\ \\ \small \bf \pink{ \frac{sin \: a(1 - cos \: a)}{ {sin}^{2} a)} \: (hence \: {sin}^{2} a - {cos}^{2} a = 1)} \\ \\ \small \bf \pink{ \therefore \: {sin}^{2}a = 1 - cos^{2} a} \\ \\ \small \bf \pink{ \frac{sin \: a(1 - cos \: a)}{sin \: a \times sin \: a} } \\ \\ \small \bf \pink{ \frac{1 - cos \: a}{sin \: a} } \\ \\ \small \bf \pink{ \frac{1}{sin \: a} - \frac{cos \: a}{sin \: a} } \\ \\ \small \bf \pink{cosec \: a - cot \: a \: \: } \\ \\ \small \bf \pink{lhs = rhs \: \: \: proved}$