Question

Prove that:
\[\dfrac{{\sin A + \cos A}}{{\sin A - \cos A}} + \dfrac{{\sin A - \cos A}}{{\sin A + \cos A}} = \dfrac{2}{{{{\sin }^2}A - {{\cos }^2}A}} = \dfrac{2}{{2{{\sin }^2}A - 1}} = \dfrac{2}{{1 - 2{{\cos }^2}A}}\]

Answer 1

Hint:
Here, we will first take into consideration the left most side of the given equation and take LCM of the fractions. Then we will simplify it using the basic algebraic and trigonometric identities to prove that it is equal to the second expression. We will then use Pythagorean identity to simplify the expression further and prove that it is equal to the third expression. Again using the same identity in the obtained expression, we will get the last expression and hence we will prove the given equation.

Formula Used:
We will use the following formulas:
1) \[{\left( {a \pm b} \right)^2} = {a^2} \pm 2ab + {b^2}\]
2) \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\]
3) Pythagorean identity: \[{\sin ^2}A + {\cos ^2}A = 1\]

Complete step by step solution:
Here, we will first consider the left hand side of the equation.
LHS \[ = \dfrac{{\sin A + \cos A}}{{\sin A - \cos A}} + \dfrac{{\sin A - \cos A}}{{\sin A + \cos A}}\]
Now, taking LCM and solving further, we get
\[ \Rightarrow \] LHS \[ = \dfrac{{{{\left( {\sin A + \cos A} \right)}^2} + {{\left( {\sin A - \cos A} \right)}^2}}}{{\left( {\sin A - \cos A} \right)\left( {\sin A + \cos A} \right)}}\]
Now, using the identities \[{\left( {a \pm b} \right)^2} = {a^2} \pm 2ab + {b^2}\] in the numerator, we get,
\[ \Rightarrow \] LHS \[ = \dfrac{{{{\sin }^2}A + 2\sin A\cos A + {{\cos }^2}A + {{\sin }^2}A - 2\sin A\cos A + {{\cos }^2}A}}{{\left( {\sin A - \cos A} \right)\left( {\sin A + \cos A} \right)}}\]
Cancelling out the same terms in the numerator and using the identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\] in the denominator, we get,
\[ \Rightarrow \] LHS \[ = \dfrac{{\left( {{{\sin }^2}A + {{\cos }^2}A} \right) + \left( {{{\sin }^2}A + {{\cos }^2}A} \right)}}{{{{\sin }^2}A - {{\cos }^2}A}}\]
Now, using the identity \[{\sin ^2}A + {\cos ^2}A = 1\] in the numerator, we get
\[ \Rightarrow \] LHS \[ = \dfrac{{1 + 1}}{{{{\sin }^2}A - {{\cos }^2}A}} = \dfrac{2}{{{{\sin }^2}A - {{\cos }^2}A}}\]
Now, from the same identity \[{\sin ^2}A + {\cos ^2}A = 1\], we can substitute the value \[{\cos ^2}A = 1 - {\sin ^2}A\] in the denominator. Therefore, we get
\[ \Rightarrow \] LHS \[ = \dfrac{2}{{{{\sin }^2}A - \left( {1 - {{\sin }^2}A} \right)}} = \dfrac{2}{{{{\sin }^2}A - 1 + {{\sin }^2}A}} = \dfrac{2}{{2{{\sin }^2}A - 1}}\]
Similarly, we can substitute the value \[{\sin ^2}A = 1 - {\cos ^2}A\] in the denominator,
\[ \Rightarrow \] LHS \[ = \dfrac{2}{{\left( {1 - {{\cos }^2}A} \right) - {{\cos }^2}A}} = \dfrac{2}{{1 - 2{{\cos }^2}A}}\]
\[ \Rightarrow \] LHS \[ = \] RHS
Therefore,
\[\dfrac{{\sin A + \cos A}}{{\sin A - \cos A}} + \dfrac{{\sin A - \cos A}}{{\sin A + \cos A}} = \dfrac{2}{{{{\sin }^2}A - {{\cos }^2}A}} = \dfrac{2}{{2{{\sin }^2}A - 1}} = \dfrac{2}{{1 - 2{{\cos }^2}A}}\]
Hence, proved

Note:
In this question, we have used trigonometry. Trigonometry is a branch of mathematics which helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine and the cosine function. In the simple terms they are written as ‘sin’, ‘cos’ and ‘tan’. Hence, trigonometry is not just a chapter to study, in fact, it is being used in everyday life.