Question

sec a -cos a× cot a + tan a = tan a sec a​

Answer 1

To Prove:-

$\\ \tt\longrightarrow( \sec a - \cos a )\times ( \cot a + \tan a) = \tan a. \sec a$

IDs Used:-

$\\ \tt\leadsto \sec \theta = \frac{1}{ \cos \theta}.. .id(1) \\ \\ \tt\leadsto \cot \theta = \frac{ \cos \theta}{ \sin \theta} ...id(2) \\ \\ \tt\leadsto \tan \theta = \frac{ \sin \theta}{ \cos \theta}. ..id(3)\\ \\ \tt\leadsto{ \sin }^{2} \theta + { \cos}^{2} \theta = 1...id(4)$

So lets siplify LHS:-

$\\ \tt\mapsto (\sec a - \cos a) \times ( \cot a + \tan a). \\ \\ \tt = ( \dfrac{1}{ \cos a} - \cos a) \times ( \frac{ \cos a }{ \sin a} + \frac{ \sin a}{ \cos a} ) . \\ \\ \rm[by \: \: usings \: \: ids(1)(2) \: \: and(3)]\\ \\ \tt = ( \frac{1 - { \cos a}^{2} }{ \cos a } ) \times ( \frac{ { \cos }^{2} a + { \sin }^{2} a }{ \sin a. \cos a} ). \\ \\ \tt = \frac{ { \sin }^{2} a }{ \cos a} \times \frac{1}{ \sin a. \cos a}. \\ \\ \rm[by \: \: using \: \: id(4)].\\ \\ \tt = \frac{ \cancel{ \sin a} \times \sin a }{ \cos a } \times \frac{1}{ \cancel{ \sin a}. \cos a } . \\ \\ \tt = \frac{ \sin a}{ \cos a } \times \frac{1}{ \cos a} . \\ \\ \tt = \tan a. \sec a = rhs \\ \\ \rm \: \: ..hence \: \: proved..$