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Question: Verify \(\dfrac{{\cos x}}{{1 - \sin x}} = \dfrac{{1 + \sin x}}{{\cos x}}\)....

Verify cos⁡x1−sin⁡x=1+sin⁡xcos⁡x\dfrac{{\cos x}}{{1 - \sin x}} = \dfrac{{1 + \sin x}}{{\cos x}}.

Explanation

Solution

Now in order to verify the above statement we should work with one side at a time and manipulate it to the other side. Using some basic trigonometric identities like we can simplify the above expression.
sin⁡2x+cos⁡2x=1 cos⁡2x=1−sin⁡2x{\sin ^2}x + {\cos ^2}x = 1 \\\ {\cos ^2}x = 1 - {\sin ^2}x
In order to verify the given expression we have to use the above identities and express our given expression in that form and thereby verify it.

Complete step by step answer:
Given, cos⁡x1−sin⁡x=1+sin⁡xcos⁡x..........................................(i)\dfrac{{\cos x}}{{1 - \sin x}} = \dfrac{{1 + \sin x}}{{\cos x}}..........................................\left( i \right)
Now in order to verify it we have to simplify either the LHS or the RHS of the equation towards RHS or the LHS of the equation respectively. Here let’s take the LHS of the equation which is:
cos⁡x1−sin⁡x\dfrac{{\cos x}}{{1 - \sin x}}
Now one of the identities that we have is cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x. So to express the above expression in terms of our identity let’s multiply the numerator and denominator with cos⁡x\cos x. Such that we get:
cos⁡x1−sin⁡x×cos⁡xcos⁡x=cos⁡2x(1−sin⁡x)cos⁡x...................(ii)\dfrac{{\cos x}}{{1 - \sin x}} \times \dfrac{{\cos x}}{{\cos x}} = \dfrac{{{{\cos }^2}x}}{{\left( {1 - \sin x} \right)\cos x}}...................\left( {ii} \right)
On applying the identity cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x we can write.
cos⁡2x(1−sin⁡x)cos⁡x=1−sin⁡2x(1−sin⁡x)cos⁡x....................(iii)\dfrac{{{{\cos }^2}x}}{{\left( {1 - \sin x} \right)\cos x}} = \dfrac{{1 - {{\sin }^2}x}}{{\left( {1 - \sin x} \right)\cos x}}....................\left( {iii} \right)
Now we know that: a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
Such that we can write:
12−sin⁡2x=(1+sin⁡x)(1−sin⁡x){1^2} - {\sin ^2}x = \left( {1 + \sin x} \right)\left( {1 - \sin x} \right)
Substituting it in (iii) we can write:
1−sin⁡2x(1−sin⁡x)cos⁡x=(1+sin⁡x)(1−sin⁡x)(1−sin⁡x)cos⁡x ∴1−sin⁡2x(1−sin⁡x)cos⁡x=1+sin⁡xcos⁡x..............(iv)\dfrac{{1 - {{\sin }^2}x}}{{\left( {1 - \sin x} \right)\cos x}} = \dfrac{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}{{\left( {1 - \sin x} \right)\cos x}} \\\ \therefore\dfrac{{1 - {{\sin }^2}x}}{{\left( {1 - \sin x} \right)\cos x}}= \dfrac{{1 + \sin x}}{{\cos x}}..............\left( {iv} \right)
Therefore from (iv) we can say that LHS =RHS.

Hence verified cos⁡x1−sin⁡x=1+sin⁡xcos⁡x\dfrac{{\cos x}}{{1 - \sin x}} = \dfrac{{1 + \sin x}}{{\cos x}}.

Note: Some other equations needed for solving these types of problem are:

\begin{array}{*{20}{l}} \Rightarrow{\sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right)} \\\ \Rightarrow{\cos \left( {2x} \right) = {{\cos }^2}\left( x \right)-{{\sin }^2}\left( x \right) = 1-2{\text{ }}{{\sin }^2}\left( x \right) = 2{\text{ }}{{\cos }^2}\left( x \right)-1} \end{array}$$ Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.