Question

Prove the trigonometric identity \[\sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}} + \sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA - 1}}} = 2\sec A\]

Answer 1

Hint:Here we rationalize each part of the equation on left side separately and then convert the trigonometric forms of \[\cos ec\theta = \dfrac{1}{{\sin \theta }}\]and substitute in the left side of the equation and using the trigonometric identity \[\cos e{c^2}\theta = 1 + {\cot ^2}\theta \]we make LHS of the equation equal to RHS of the equation.

Complete step-by-step answer:
First we rationalize each part of the equation on LHS.
We rationalize the term \[\sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}} \] by multiplying both numerator and denominator by \[\sqrt {\cos ecA - 1} \]
So, \[\sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}} \times \sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA - 1}}} = \sqrt {\dfrac{{(\cos ecA - 1) \times (\cos ecA - 1)}}{{(\cos ecA + 1) \times (\cos ecA - 1)}}} \]
                                                            \[ = \sqrt {\dfrac{{{{(\cos ecA - 1)}^2}}}{{{{(\cos ecA)}^2} - {1^2}}}} \] { since \[(a - b)(a + b) = {a^2} - {b^2}\]}
Substitute the value \[\cos e{c^2}\theta = 1 + {\cot ^2}\theta \] in the denominator
\[\sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}} \times \sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA - 1}}} = \sqrt {\dfrac{{{{(\cos ecA - 1)}^2}}}{{1 + {{\cot }^2}A - {1^2}}}} \]
     \[
   = \sqrt {\dfrac{{{{(\cos ecA - 1)}^2}}}{{{{\cot }^2}A}}} \\
   = \sqrt {{{\left( {\dfrac{{\cos ecA - 1}}{{\cot A}}} \right)}^2}} \\
 \] { taking the squares common }
 Now we can cancel out the square root by square of the whole value.
\[\sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}} \times \sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA - 1}}} = \dfrac{{\cos ecA - 1}}{{\cot A}}\]
Similarly we rationalize the second part of LHS.
We rationalize the term \[\sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA - 1}}} \] by multiplying both numerator and denominator by \[\sqrt {\cos ecA + 1} \]
So, \[\sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA - 1}}} \times \sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA + 1}}} = \sqrt {\dfrac{{(\cos ecA + 1) \times (\cos ecA + 1)}}{{(\cos ecA - 1) \times (\cos ecA + 1)}}} \]
                                                            \[ = \sqrt {\dfrac{{{{(\cos ecA + 1)}^2}}}{{{{(\cos ecA)}^2} - {1^2}}}} \] { since \[(a - b)(a + b) = {a^2} - {b^2}\]}
Substitute the value \[\cos e{c^2}\theta = 1 + {\cot ^2}\theta \] in the denominator
\[\sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA - 1}}} \times \sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA + 1}}} = \sqrt {\dfrac{{{{(\cos ecA + 1)}^2}}}{{1 + {{\cot }^2}A - {1^2}}}} \]
     \[
   = \sqrt {\dfrac{{{{(\cos ecA + 1)}^2}}}{{{{\cot }^2}A}}} \\
   = \sqrt {{{\left( {\dfrac{{\cos ecA + 1}}{{\cot A}}} \right)}^2}} \\
 \] { taking the squares common }
 Now we can cancel out the square root by square of the whole value.
\[\sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA - 1}}} \times \sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA + 1}}} = \dfrac{{\cos ecA + 1}}{{\cot A}}\]
Now we substitute both the values in LHS of the equation.
\[\sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}} + \sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA - 1}}} = \dfrac{{\cos ecA - 1}}{{\cot A}} + \dfrac{{\cos ecA + 1}}{{\cot A}}\]
Taking LCM on the RHS of the equation.
\[\sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}} + \sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA - 1}}} = \dfrac{{\cos ecA - 1 + \cos ecA + 1}}{{\cot A}}\]
                                                       \[ = \dfrac{{2\cos ecA}}{{\cot A}}\]
Substituting the values of \[\cos ecA = \dfrac{1}{{\sin A}}\] and \[\cot A = \dfrac{{\cos A}}{{\sin A}}\]
\[\sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}} + \sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA - 1}}} = \dfrac{{2\dfrac{1}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}}}}\]
                                                       \[
   = \dfrac{2}{{\sin A}} \times \dfrac{{\sin A}}{{\cos A}} \\
   = \dfrac{2}{{\cos A}} \\
 \] { Cancel the same terms }
Now substitute the value of \[\dfrac{1}{{\cos A}} = \sec A\]
\[\sqrt {\dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}} + \sqrt {\dfrac{{\cos ecA + 1}}{{\cos ecA - 1}}} = 2\sec A\]
Hence proved.

Note:Students are likely to make mistakes solving the question without rationalizing the terms under the square root and the terms in fraction form which makes the solution more complicated. Keep in mind we always rationalize first and then proceed with the solution. Also always try to convert the term under the square root as a square term so as to cancel out the root.