Question

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Prove that (sin alpha +cos alpha) (tan alpha+cot alpha) =sec alpha + cosec alpha​

Answer 1

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

To prove :

$( \sin( \alpha ) + \cos( \alpha ) )( \tan( \alpha ) + \cot( \alpha ) ) = \sec( \alpha ) + \csc( \alpha )$

LHS

$( \sin( \alpha ) + \cos( \alpha ) )( \tan( \alpha ) + \cot( \alpha ) ) \\ \\ \leadsto \: ( \sin( \alpha ) + \cos( \alpha ) )( \frac{ \sin( \alpha ) }{ \cos( \alpha ) } + \frac{ \cos( \alpha ) }{ \sin( \alpha ) } \\ \\ \leadsto \: ( \sin( \alpha ) + \cos( \alpha ) )( \frac{ \sin ^{2} ( \alpha ) + \cos ^{2} ( \alpha ) }{ \cos( \alpha ) \sin( \alpha ) } ) \\ \\ \leadsto \: ( \sin( \alpha ) + \cos( \alpha ) )( \frac{1}{ \cos( \alpha ) \sin( \alpha ) } ) \\ \\ \leadsto \: \frac{ \sin( \alpha ) }{ \cos( \alpha ) \sin( \alpha ) } + \frac{ \cos( \alpha ) }{ \cos( \alpha ) \sin( \alpha ) } \\ \\ \leadsto \: \cancel \frac{ \sin( \alpha ) }{ \sin( \alpha ) \: \cos( \alpha ) } + \cancel \frac{ \cos( \alpha ) }{ \cos( \alpha ) \sin( \alpha ) } \\ \\ \leadsto \frac{1}{ \cos( \alpha ) } + \frac{1}{ \sin( \alpha ) } \\ \\ \leadsto \: \sec( \alpha ) + \csc( \alpha )$

LHS = RHS

$\huge\boxed{ \green{ \bold{proved}}}$

Formula used

◆sin^2@+cos^2@=1

◆ tan@= sin@/cos@

◆cot@= cos@/sin@

Answer 2

(sin +cos )(tan+cot)

=(sin+cos)(sin/cos+cos/sin)

=(sin+cos)(sin^2+cos^2/sin cos)

=(sin+cos)(1/sin cos)

=sin/sin cos + cos/ sin cos

= 1/cos+1/sin

=sec + csc

LHS = RHS