Question

Prove that $\dfrac{{\cos A}}{{1 + \sin A}} = \tan \left( {\dfrac{\pi }{4} - \dfrac{A}{2}} \right)$

Answer 1

Hint : Here, we need to prove the given equation. And for this, we will solve the LHS part and compare it with the RHS part. We will also use some trigonometric ratios identities like, $1 = {\cos ^2}A + {\sin ^2}A$ , $\cos 2A = {\cos ^2}A - {\sin ^2}A$ , $\sin 2A = 2\sin A\cos A$ and $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$ . Also we will use some formulas too and they are, ${a^2} - {b^2} = (a - b)(a + b)$ and ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$ . We must also know the value of $\tan \left( {\dfrac{\pi }{4}} \right) = 1$ . We will use all these identities to get the final output.

Complete step-by-step answer :
Given that,
$\dfrac{{\cos A}}{{1 + \sin A}} = \tan \left( {\dfrac{\pi }{4} - \dfrac{A}{2}} \right)$
Here, we will first prove the LHS part and then will compare it with the RHS part.
LHS
$= \dfrac{{\cos A}}{{1 + \sin A}}$
Multiply and divide by 2 in both the trigonometric ratios, we will get,
$= \dfrac{{\cos 2\left( {\dfrac{A}{2}} \right)}}{{1 + \sin 2\left( {\dfrac{A}{2}} \right)}}$
We know that, $\cos 2A = {\cos ^2}A - {\sin ^2}A$ and so applying this identity, we will get,
$= \dfrac{{{{\cos }^2}\left( {\dfrac{A}{2}} \right) - {{\sin }^2}\left( {\dfrac{A}{2}} \right)}}{{1 + \sin 2\left( {\dfrac{A}{2}} \right)}}$
We know this identity that, $1 = {\cos ^2}A + {\sin ^2}A$ and so applying this, we will get,
$= \dfrac{{{{\cos }^2}\left( {\dfrac{A}{2}} \right) - {{\sin }^2}\left( {\dfrac{A}{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{A}{2}} \right) + {{\sin }^2}\left( {\dfrac{A}{2}} \right) + \sin 2\left( {\dfrac{A}{2}} \right)}}$
We know that, $\sin 2A = 2\sin A\cos A$ and so applying this, we will get,
$= \dfrac{{{{\cos }^2}\left( {\dfrac{A}{2}} \right) - {{\sin }^2}\left( {\dfrac{A}{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{A}{2}} \right) + {{\sin }^2}\left( {\dfrac{A}{2}} \right) + 2\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{A}{2}} \right)}}$
We also know this formula, ${a^2} - {b^2} = (a - b)(a + b)$ and here, we have rearrange the denominator, we will get,
$= \dfrac{{\left[ {\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)} \right]\left[ {\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)} \right]}}{{{{\cos }^2}\left( {\dfrac{A}{2}} \right) + 2\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{A}{2}} \right) + {{\sin }^2}\left( {\dfrac{A}{2}} \right)}}$
We will use this formula, ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$ and so applying this, we will get,
$= \dfrac{{\left[ {\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)} \right]\left[ {\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)} \right]}}{{{{\left[ {\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)} \right]}^2}}}$
Opening the brackets of the denominator, we will get,
$= \dfrac{{\left[ {\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)} \right]\left[ {\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)} \right]}}{{\left[ {\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)} \right]\left[ {\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)} \right]}}$
$= \dfrac{{\left[ {\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)} \right]}}{{\left[ {\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)} \right]}}$ ---------- (1)
Dividing both the numerator and denominator by$\cos \left( {\dfrac{A}{2}} \right)$ , we will get,
$= \dfrac{{\dfrac{{\cos \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}} - \dfrac{{\sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}}}}{{\dfrac{{\cos \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}} + \dfrac{{\sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}}}}$
We know that, $\tan A = \dfrac{{\sin A}}{{\cos A}}$ , we will get,
$= \dfrac{{1 - \tan \left( {\dfrac{A}{2}} \right)}}{{1 + \tan \left( {\dfrac{A}{2}} \right)}}$
We know the value of $\tan \left( {\dfrac{\pi }{4}} \right) = 1$ and so applying this, we will get,
$= \dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) - \tan \left( {\dfrac{A}{2}} \right)}}{{1 + \tan \left( {\dfrac{\pi }{4}} \right)\tan \left( {\dfrac{A}{2}} \right)}}$
We will use this identity $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$ and using this, we will get,
$= \tan \left( {\dfrac{\pi }{4} - \dfrac{A}{2}} \right)$
=RHS

Note : After solving the LHS part till equation (1), we can start solving the RHS part too…
RHS
$= \tan \left( {\dfrac{\pi }{4} - \dfrac{A}{2}} \right)$
$= \dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) - \tan \left( {\dfrac{A}{2}} \right)}}{{1 + \tan \left( {\dfrac{\pi }{4}} \right)\tan \left( {\dfrac{A}{2}} \right)}}$
$= \dfrac{{1 - \tan \left( {\dfrac{A}{2}} \right)}}{{1 + \tan \left( {\dfrac{A}{2}} \right)}}$
$= \dfrac{{1 - \dfrac{{\sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}}}}{{1 + \dfrac{{\sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}}}}$
$= \dfrac{{\dfrac{{\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}}}}{{\dfrac{{\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right)}}}}$
$= \dfrac{{\cos \left( {\dfrac{A}{2}} \right) - \sin \left( {\dfrac{A}{2}} \right)}}{{\cos \left( {\dfrac{A}{2}} \right) + \sin \left( {\dfrac{A}{2}} \right)}}$
= LHS till equation (1)
Thus, LHS = RHS.
Hence, it is proved $\dfrac{{\cos A}}{{1 + \sin A}} = \tan \left( {\dfrac{\pi }{4} - \dfrac{A}{2}} \right)$ .