Question

Prove :
 \[\dfrac{{1 + cosx + sinx}}{{1 + cosx - sinx}} = \dfrac{{1 + sinx}}{{cosx}}\;\]

Answer 1

Hint: To prove the given trigonometric expression start solving from LHS. First divide each term of numerator and denominator by \[\;cosx\] then apply the trigonometric formulas as required and solve LHS we will get the RHS.

Complete step-by-step answer:
LHS is given as
 \[\dfrac{{1 + cosx + sinx}}{{1 + cosx - sinx}}\]
 Dividing each terms of numerators and denominator by \[\;cosx\]
We get,
 \[\dfrac{{secx + 1 + tanx}}{{secx + 1 - tanx}}\]
now write \[1 = se{c^2}x - ta{n^2}x\] in the above equation
 \[\dfrac{{secx + tanx + se{c^2}x - ta{n^2}x}}{{secx + 1 - tanx}}\]
now break \[se{c^2}x - ta{n^2}x\] ​ into \[\left( {secx - tanx} \right)\left( {secx + tanx} \right)\]
 \[\dfrac{{secx + tanx + \left( {secx - tanx} \right)\left( {secx + tanx} \right)}}{{secx + 1 - tanx}}\]
now take \[secx + tanx\] as common
 \[\dfrac{{secx + tanx\left( {secx - tanx + 1} \right)}}{{secx + 1 - tanx}}\]
cancelling the like terms we get,
 \[secx + tanx\]
now write all the trigonometric ratios in terms of $ \cos x $ and \[\;sinx\]
we get,
 \[\dfrac{1}{{cosx}} + \dfrac{{sinx}}{{cosx}} = \dfrac{{1 + sinx}}{{cosx}}\]
Hence \[LHS = RHS\]

Note: Instead of dividing each term of numerator and denominator by $ \cos x $ we can also divide by \[\;sinx\] each term of numerator and denominator we will get the same result. Students can take this as an example to try out the other method.