Question

Prove that 1/sec a -tan a= secA tanA

Answer 1

Answer:

Please see the attachment

Answer 2

$\boxed{\underline{{ \huge{ \mathrm{A}}} \mathrm{NSWER :-}}}$

$\red{ \tt{ \: \frac{1}{sec \: A- tan \: A} = sec \: A.tan \: A }} \\ \\ : \longrightarrow \blue{ \tt{ \: \frac{1}{ \frac{1}{cos \: A} - \frac{sin \: A}{cos \:A } } }} \\ \\ \\ : \longrightarrow \blue{ \tt{ \: \dfrac{1}{ \frac{cos \: A - sin \: A.cos \:A }{ {cos}^{2} A} } }} \\ \\ \tt{ \: take \: cos \: A \: common ;} \\ \\ : \longrightarrow \blue{ \tt{ \: \frac{1}{ \frac{ {\cancel{cos \:A}}(sin \: A + 1)}{ {cos}^{ \cancel{2}} A} } }} \\ \\ \\ : \longrightarrow \blue{ \tt{ \: \frac{1}{cos \:A }.(sin \: A + 1)}} \\ \\ \\ : \longrightarrow \blue{ \tt{ \: sec \: A.tan \:A }} \\ \\ \therefore \tt{ \: L.H.S = R.H.S \: \: \: \: \: \: \: \: \: (proved)}$

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