Question

( secθ-cosθ)( cotθ+tanθ) = tanθ secθ ,Prove it

Answer 1

We have,

$( \sec( \theta) - \cos( \theta) ) \times ( \cot( \theta) + \tan( \theta) ) \\ = > ( \frac{1}{ \cos(\theta) } - \cos(\theta) )( \frac{ \cos(\theta) }{ \sin(\theta )} + \frac{ \sin( \theta) }{ \cos( \theta) } ) \\ = > (\frac{1 - {cos}^{2} ( \theta)}{ \cos( \theta) } )( \frac{ { \cos}^{2}( \theta) + { \sin }^{2} ( \theta)}{ \sin( \theta) \cos( \theta) } ) \\ = > \frac{ { \sin }^{2}( \theta) }{ \cos( \theta) } \times \frac{1}{ \sin( \theta) \cos( \theta) } \\ = > \frac{ \sin( \theta) }{ \cos( \theta) } \times \frac{1}{ \cos( \theta) } \\ = > \tan( \theta) \times \sec( \theta)$
Hence Proved

Answer 2

SOLUTION:-
GIVEN BY:- ( secθ-cosθ)( cotθ+tanθ) = tanθ secθ.
L.H.S =>

$( secθ-cosθ)( cotθ+tanθ) \\ = (\frac{1}{cosθ} - cosθ)( \frac{cosθ}{sinθ} + \frac{sinθ}{cosθ} ) \\ = ( \frac{1 - {(cosθ)}^{2} }{cosθ} )( \frac{cosθ}{sinθ} + \frac{sinθ}{cosθ} ) \\ ( \frac{ {(sinθ)}^{2} }{cosθ} )( \frac{cosθ}{sinθ} + \frac{sinθ}{cosθ} ) \\ = ( \frac{ {(sinθ)}^{2} }{cosθ} )( \frac{ {(cosθ )}^{2} + {(sinθ )}^{2} }{ sinθ.cosθ } \\ = \frac{sinθ}{ {(cosθ)}^{2} } \\ = \frac{ sinθ}{cosθ} . \frac{1}{cosθ} \\ = \: tanθ.secθ$
here

R.HS = L.H.S


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