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.