Question

prove tan + 1/tan = sec.cosec​

Answer 1

Answer:

Please see the attachment

Answer 2

$\huge \red{ \mathfrak{solution}}$

$\tan( \alpha ) + \frac{1}{ \tan( \alpha ) } = \sec( \alpha ) \csc( \alpha )$

LHS

$\tan( \alpha ) + \frac{1}{ \tan( \alpha ) }$

$\rightarrow \: \frac{ \sin( \alpha ) }{ \cos( \alpha ) } + \frac{1}{ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } \\ \\ \rightarrow \: \frac{ \sin( \alpha ) }{ \cos( \alpha ) } + \frac{ \cos( \alpha ) }{ \sin( \alpha ) } \\ \\ \rightarrow \: \frac{ { \sin }^{2} (\alpha )+ \cos ^{2} ( \alpha ) }{ \cos( \alpha ) \sin( \alpha ) } \\ \\ \rightarrow \: \frac{1}{ \cos( \alpha ) \sin( \alpha ) } \\ \\ \rightarrow \: \sec( \alpha ) \csc( \alpha )$

LHS= RHS

$\huge \green{ \boxed{ \mathfrak{poved}}}$

Formula used :

• tan @= sin@/cos@

• sin^2@+cos^2@ =1

• cos@=1/sec@

• sin@=1/ csc@