Question

(secx-cosx)(cotx+tanx)=tanx*secx

Answer 1

Question ;

$( \sec(x) - \cos(x) )( \cot(x) + \tan( x ) ) = \tan(x) \times \sec(x)$

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LH S :

$( \sec(x) - \cos(x) )( \cot(x) + \tan(x) )$

$= ( \frac{1}{ \cos(x) } - \cos(x) )( \frac{ \cos(x) }{ \sin(x) } + \frac{ \sin(x) }{ \cos(x) } )$

$= ( \frac{1 - \cos {}^{2} (x) }{ \cos(x) } )( \frac{ \sin {}^{2} (x) + \cos {}^{2} (x) }{ \sin(x) \times \cos(x) } )$

$= (\frac{ \sin {}^{2} ( x ) }{ \cos(x) }) ( \frac{1}{ \cos(x) \times \sin(x) } )$

$= \frac{ \sin(x) }{ \cos(x) } \times \frac{1}{ \cos(x) }$

$= \tan(x) \times \sec(x)$

= RHS

Answer 2

Step-by-step explanation:

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