Question

Prove: $\dfrac{{\cos A}}{{\left( {1 + \sin A} \right)}} + \dfrac{{\left( {1 + \sin A} \right)}}{{\cos A}} = 2\sec A$

Answer 1

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 ${\sin ^2}A + {\cos ^2}A = 1$ . 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 must know how to take LCM of two terms and simplify the addition of two rational numbers in order to solve the question.

Complete step-by-step answer :
In the given problem, we have to prove a trigonometric equality that can be further used in many questions and problems as a direct result and has wide ranging applications. 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. $= \dfrac{{\cos A}}{{\left( {1 + \sin A} \right)}} + \dfrac{{\left( {1 + \sin A} \right)}}{{\cos A}}$
Now, we multiply the numerator and denominator of the first rational term by $\left( {1 - \sin x} \right)$ in order to simplify the rational trigonometric expression. So, we get,
$\Rightarrow \dfrac{{\cos A}}{{\left( {1 + \sin A} \right)}} \times \dfrac{{\left( {1 - \sin A} \right)}}{{\left( {1 - \sin A} \right)}} + \dfrac{{\left( {1 + \sin A} \right)}}{{\cos A}}$
$\Rightarrow \dfrac{{\cos A\left( {1 - \sin A} \right)}}{{\left( {1 + \sin A} \right)\left( {1 - \sin A} \right)}} + \dfrac{{\left( {1 + \sin A} \right)}}{{\cos A}}$
Now, simplifying the denominator by using the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ , we get,
$\Rightarrow \dfrac{{\cos A\left( {1 - \sin A} \right)}}{{{1^2} - {{\sin }^2}A}} + \dfrac{{\left( {1 + \sin A} \right)}}{{\cos A}}$
Now, applying the trigonometric identity ${\cos ^2}A + {\sin ^2}A = 1$, we get,
$\Rightarrow \dfrac{{\cos A\left( {1 - \sin A} \right)}}{{{{\cos }^2}A}} + \dfrac{{\left( {1 + \sin A} \right)}}{{\cos A}}$
Cancelling the common factors in numerator and denominator, we get,
$\Rightarrow \dfrac{{\left( {1 - \sin A} \right)}}{{\cos A}} + \dfrac{{\left( {1 + \sin A} \right)}}{{\cos A}}$
Now, the denominator of both the fractions are the same. So, we can directly add up the numerators.
$\Rightarrow \dfrac{{1 - \sin A + 1 + \sin A}}{{\cos A}}$
$\Rightarrow \dfrac{2}{{\cos A}}$
Now, we know that cosine and secant are reciprocal trigonometric functions. So, we have,
$\Rightarrow 2\sec A$
Now, L.H.S $= 2\sec A$
As the left side of the equation is equal to the right side of the equation, we have,
$\dfrac{{\cos A}}{{\left( {1 + \sin A} \right)}} + \dfrac{{\left( {1 + \sin A} \right)}}{{\cos A}} = 2\sec A$
Hence, Proved.

Note: Given 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.