Question

Show that: $\dfrac{{\sin A}}{{\sec A + \tan A - 1}} + \dfrac{{\cos A}}{{\cos ecA + \cot A - 1}} = 1$

Answer 1

Hint: The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $\sec x = \dfrac{1}{{\cos x}}$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$. Basic algebraic rules and trigonometric identities are to be kept in mind while simplifying the given problem and proving the result given to us. We will first convert all the trigonometric functions into sine and cosine and then simplify the expression by cancelling the common factors in numerator and denominator.

Complete answer: In the given problem, we have to prove a trigonometric equality. For proving the desired result, we need to have a good grip over the basic trigonometric formulae and identities.
Now, we need to make the left and right sides of the equation equal.
L.H.S. $ = \dfrac{{\sin A}}{{\sec A + \tan A - 1}} + \dfrac{{\cos A}}{{\cos ecA + \cot A - 1}}$
Now, we convert the trigonometric functions into sine and cosine using some basic trigonometric formulae $\sec x = \dfrac{1}{{\cos x}}$, $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$. So, we get,
$ = \dfrac{{\sin A}}{{\dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}} - 1}} + \dfrac{{\cos A}}{{\dfrac{1}{{\sin A}} + \dfrac{{\cos A}}{{\sin A}} - 1}}$
Now, we take the LCM of rational expressions in denominator, we get,
$ = \dfrac{{\sin A}}{{\dfrac{{1 + \sin A - \cos A}}{{\cos A}}}} + \dfrac{{\cos A}}{{\dfrac{{1 + \cos A - \sin A}}{{\sin A}}}}$
Simplifying the equation, we get,
$ = \dfrac{{\sin A\cos A}}{{1 + \left( {\sin A - \cos A} \right)}} + \dfrac{{\sin A\cos A}}{{1 - \left( {\sin A - \cos A} \right)}}$
Now, multiplying the numerator and denominator by the same number, we get,
$ = \dfrac{{\sin A\cos A}}{{1 + \left( {\sin A - \cos A} \right)}} \times \left[ {\dfrac{{1 - \left( {\sin A - \cos A} \right)}}{{1 - \left( {\sin A - \cos A} \right)}}} \right] + \dfrac{{\sin A\cos A}}{{1 - \left( {\sin A - \cos A} \right)}} \times \left[ {\dfrac{{1 + \left( {\sin A - \cos A} \right)}}{{1 + \left( {\sin A - \cos A} \right)}}} \right]$
Now, using the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$,
$ = \dfrac{{\sin A\cos A\left( {1 - \sin A + \cos A} \right)}}{{1 - {{\left( {\sin A - \cos A} \right)}^2}}} + \dfrac{{\sin A\cos A\left( {1 + \sin A - \cos A} \right)}}{{1 - {{\left( {\sin A - \cos A} \right)}^2}}}$
Now, adding the numerators as the denominators, we get,
$ = \dfrac{{\sin A\cos A\left( {1 - \sin A + \cos A} \right) + \sin A\cos A\left( {1 + \sin A - \cos A} \right)}}{{1 - \left( {{{\sin }^2}A + {{\cos }^2}A - 2\sin A\cos A} \right)}}$
Opening the brackets and cancelling the like terms with opposite signs, we get,
$ = \dfrac{{\sin A\cos A - {{\sin }^2}A\cos A + \sin A{{\cos }^2}A + \sin A\cos A + {{\sin }^2}A\cos A - \sin A{{\cos }^2}A}}{{1 - \left( {1 - 2\sin A\cos A} \right)}}$
$ = \dfrac{{\sin A\cos A + \sin A\cos A}}{{1 - 1 + 2\sin A\cos A}}$
Cancelling the common factors in numerator and denominator, we get,
$ = \dfrac{{2\sin A\cos A}}{{2\sin A\cos A}}$
$ = 1$
R.H.S. $ = 1$
As the left side of the equation is equal to the right side of the equation, we have,
$\dfrac{{\sin A}}{{\sec A + \tan A - 1}} + \dfrac{{\cos A}}{{\cos ecA + \cot A - 1}} = 1$
Hence, Proved.

Note:
The problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. We must also express rational functions in simplest forms by cancelling the common factors in numerator and denominator.