Question

Prove that \[\dfrac{{\sin A}}{{1 + \cos A}} + \dfrac{{1 + \cos A}}{{\sin A}} = 2\cos ecA\]

Answer 1

Hint: It is a simple trigonometry question first do the LCM and then use the formula \[{\sin ^2}A + {\cos ^2}A = 1\] to reach the final answer. Also it must be remembered that \[\cos ec\theta = \dfrac{1}{{\sin \theta }}\& \sec \theta = \dfrac{1}{{\cos \theta }}\]

Complete step-by-step answer:
We will start from LHS and then try to go towards the RHS part just like any conventional proof.
In the LHS we are given that
\[\dfrac{{\sin A}}{{1 + \cos A}} + \dfrac{{1 + \cos A}}{{\sin A}}\]
Let us first do the LCM and see what we can get
\[\begin{array}{l}
 = \dfrac{{\sin A \times \sin A + (1 + \cos A)(1 + \cos A)}}{{\sin A(1 + \cos A)}}\\
 = \dfrac{{{{\sin }^2}A + {{\left( {1 + \cos A} \right)}^2}}}{{\sin A(1 + \cos A)}}\\
 = \dfrac{{{{\sin }^2}A + 1 + {{\cos }^2}A + 2\cos A}}{{\sin A(1 + \cos A)}}
\end{array}\]
Let us put \[{{{\sin }^2}A + {{\cos }^2}A = 1}\] in the above step and we will try to solve it that way by taking command in numerator and denominator and cancelling them out
\[\begin{array}{l}
 = \dfrac{{1 + 1 + 2\cos A}}{{\sin A(1 + \cos A)}}\\
 = \dfrac{{2 + 2\cos A}}{{\sin A(1 + \cos A)}}\\
 = \dfrac{{2(1 + \cos A)}}{{\sin A(1 + \cos A)}}\\
 = \dfrac{2}{{\sin A}}\\
 = 2\cos ecA
\end{array}\]
So from here it is clear that LHS=RHS hence proved.

Note: It must be noted that \[\dfrac{1}{{\sin A}} = \cos ecA\] also many students usually put \[{{{\sin }^2}A + {{\cos }^2}A}\] in place of 1 which makes the calculation much longer.Just like \[{{{\sin }^2}A + {{\cos }^2}A = 1}\] we also have \[{\sec ^2}A - {\tan ^2}A = 1\& \cos e{c^2}A - {\cot ^2}A = 1\] Also note that \[\sin \theta = \dfrac{p}{h},\cos \theta = \dfrac{b}{h},\tan \theta = \dfrac{p}{b},\cos ec\theta = \dfrac{h}{p},\sec \theta = \dfrac{h}{b},\cot \theta = \dfrac{b}{p}\] where p, b, h represents perpendicular, base and height of a right angled triangle respectively.