Question

How do you prove that $\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) + \left( {\dfrac{{1 + \sin x}}{{\cos x}}} \right) = 2\sec x?$

Answer 1

Hint: Consider the LHS of the given question. Start by taking the LCM of the two terms. Multiply and simplify the expression using the identity ${\cos ^2}x + {\sin ^2}x = 1$. The final expression can be obtained by using the conversion $\dfrac{1}{{\cos x}} = \sec x$.

Complete Step by Step Solution:
We need to prove $\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) + \left( {\dfrac{{1 + \sin x}}{{\cos x}}} \right) = 2\sec x$
To prove the expression, we simply need to prove LHS=RHS.
LHS: $\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) + \left( {\dfrac{{1 + \sin x}}{{\cos x}}} \right)$
Taking the LCM of the terms
$ = \dfrac{{\left( {\cos x} \right)\left( {\cos x} \right) + \left( {1 + \sin x} \right)\left( {1 + \sin x} \right)}}{{\left( {1 + \sin x} \right)\left( {\cos x} \right)}}$
Multiplying the terms in the numerator
$ = \dfrac{{{{\cos }^2}x + \left( {1 + \sin x + \sin x + {{\sin }^2}x} \right)}}{{\left( {1 + \sin x} \right)\left( {\cos x} \right)}}$
$ = \dfrac{{{{\cos }^2}x + 1 + 2\sin x + {{\sin }^2}x}}{{(1 + \sin x)(\cos x)}}$
Using the identity ${\cos ^2}x + {\sin ^2}x = 1$ the expression can be simplified as
$ = \dfrac{{1 + 1 + 2\sin x}}{{(1 + \sin x)(\cos x)}}$
$ = \dfrac{{2 + 2\sin x}}{{(1 + \sin x)(\cos x)}}$
Taking 2 common from both the terms in the numerator the equation becomes
$ = \dfrac{{2(1 + \sin x)}}{{(1 + \sin x)(\cos x)}}$
Now, we have $(1 + \sin x)$ common in both the numerator and denominator. Cancelling the common terms
$ = \dfrac{2}{{\cos x}}$
Using the conversion $\dfrac{1}{{\cos x}} = \sec x$ the expression can be re-written as
$ = 2\sec x$ = RHS
LHS=RHS. Hence, Proved.

Note:
Alternate Method:
We need to prove $\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) + \left( {\dfrac{{1 + \sin x}}{{\cos x}}} \right) = 2\sec x$
To prove the expression, we simply need to prove LHS=RHS.
LHS: $\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) + \left( {\dfrac{{1 + \sin x}}{{\cos x}}} \right)$
Multiplying and dividing each term by $(1 - \sin x)$
$ = \dfrac{{\cos x\left( {1 - \sin x} \right)}}{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}} + \dfrac{{(1 + \sin x)(1 - \sin x)}}{{\cos x(1 - \sin x)}}$
Using the identity $(a + b)(a - b) = {a^2} - {b^2}$ to simplify the denominator of the first term and the denominator of the second term
$ = \dfrac{{\cos x(1 - \sin x)}}{{1 - {{\sin }^2}x}} + \dfrac{{1 - {{\sin }^2}x}}{{\cos x(1 - \sin x)}}$
We can use the identity $1 - {\sin ^2}x = {\cos ^2}x$ to write the equation as
$ = \dfrac{{\cos x(1 - \sin x)}}{{{{\cos }^2}x}} + \dfrac{{{{\cos }^2}x}}{{\cos x(1 - \sin x)}}$
Cancelling the common terms in the numerator and denominator of both the terms
$ = \dfrac{{(1 - \sin x)}}{{\cos x}} + \dfrac{{\cos x}}{{(1 - \sin x)}}$
Taking the LCM of the terms and multiplying
$ = \dfrac{{{{(1 - \sin x)}^2} + {{\cos }^2}x}}{{\cos x(1 - \sin x)}}$
Using the identity ${(a - b)^2} = {a^2} - 2ab + {b^2}$ , we can expand
$ = \dfrac{{1 - 2\sin x + {{\sin }^2}x + {{\cos }^2}x}}{{\cos x(1 - \sin x)}}$
Using the identity ${\cos ^2}x + {\sin ^2}x = 1$ the expression can be simplified as
$ = \dfrac{{1 - 2\sin x + 1}}{{\cos x(1 - \sin x)}}$
$ = \dfrac{{2 - 2\sin x}}{{\cos x(1 - \sin x)}}$
Taking 2 common from both the terms in the numerator the equation becomes
$ = \dfrac{{2(1 - \sin x)}}{{\cos x(1 - \sin x)}}$
Now, we have $(1 - \sin x)$ common in both the numerator and denominator. Cancelling the common terms
$ = \dfrac{2}{{\cos x}}$
Using the conversion $\dfrac{1}{{\cos x}} = \sec x$ the expression can be re-written as
$ = 2\sec x$ = RHS
LHS=RHS. Hence, Proved.