Question

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

Answer 1

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

= \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.