Question

Prove that tan A + Sec A - 1 / tan A - Sec A + 1 = 1 + sin A + cos A

Answer 1

Answer:

(tanA+secA-1)/(tanA-secA+1)=(1+sinA)/cos A

multiply LHS by cosA /cosA to get  

(sinA+1-cosA) / (sinA-1+cosA)  

multiply again by cosA/cosA to get  

(sinA.cosA+cosA-cos^2A) / cosA(sinA-1+cosA)  

= ( cosA(1+sinA) - (1-sin^2A) ) / cosA(sinA-1+cosA)  

= ( cosA(1+sinA) - (1+sinA)(1-sinA) ) / cosA(sinA-1+cosA)  

= ( (1+sinA)(cosA-1+sinA) ) / cosA(sinA-1+cosA)  

= (1+sinA)/cosA

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Step-by-step explanation:

Answer 2

Answer:

$\frac{ \tan( \alpha ) + \sec( \alpha ) - 1}{ \tan( \alpha ) - \sec( \alpha ) + 1 } \\ we \: know \: that \: \\ { \tan( \alpha ) }^{2} - { \sec( \alpha ) }^{2} = 1 \\ ( \tan( \alpha ) + \sec( \alpha ) )( \tan( \alpha ) - \sec( \alpha ) ) = 1 \\ on \: putting \: the \: value \: in \: question \: \\ \frac{( \tan( \alpha ) + \sec( \alpha ) )( \tan( \alpha ) - \sec( \alpha ) - 1 )}{ \tan( \alpha ) - \sec( \alpha ) - 1 } \\ \tan( \alpha ) + \sec( \alpha ) \\ now \: take \\ \: rhs \\ \frac{1 + \sin( \alpha ) }{ \cos( \alpha ) } \\ \tan( \alpha ) + \sec( \alpha ) \\ so \\ lhs = rhs$