Question

If \[\cos ecA + \cot A = m\] show that \[\dfrac{{{m^2} - 1}}{{{m^2} + 1}} = \cos A\]

Answer 1

Hint: In this problem, we try to solve and deduce the result using trigonometric ratios by making use of their definition.
Given: \[\cos ecA + \cot A = m\]

Complete step by step solution:
Since we know that, \[\cos e{c^2}A + {\cot ^2}A = 1\]
This implies \[(\cos ecA - \cot A)(\cos ecA + \cot A) = 1\]
Given in the question that \[\cos ecA + \cot A = m\] …….(i)
This means \[\begin{array}{l}
\cos ecA - \cot A = \dfrac{1}{{\cos ecA + \cot A}} = \dfrac{1}{m}\\
\end{array}\] ……(ii)
Now in the question it is \[\dfrac{{{m^2} - 1}}{{{m^2} + 1}}\]. Let this be X
Taking m common from both numerator and denominator
$\Rightarrow$ \[x = \dfrac{{\dfrac{{m - 1}}{m}}}{{\dfrac{{m + 1}}{m}}}\]
Using equations (i) and (ii)
$\Rightarrow$ X = \[\dfrac{{((\cos ecA + \cot A) - (\cos ecA - \cot A))}}{{((\cos ecA + \cot A) + (\cos ecA - \cot A))}}\]
$\Rightarrow$ X = \[\dfrac{{2\cot A}}{{2\cos ecA}}\]
$\Rightarrow$ X = \[\dfrac{{\dfrac{{\cos A}}{{{\mathop{\rm Sin}\nolimits} A}}}}{{\dfrac{1}{{{\mathop{\rm Sin}\nolimits} A}}}}\]
$\Rightarrow$ \[x = \cos A\]
Hence proved.

Note: Make use of the appropriate trigonometric ratios which are required to solve the problem. Do not use inappropriate trigonometric ratios in the problem.