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Question: How do you prove that \(\dfrac{{\cos x}}{{1 - \sin x}} - \dfrac{1}{{\cos x}} = \tan x\) ?...

How do you prove that cos⁡x1−sin⁡x−1cos⁡x=tan⁡x\dfrac{{\cos x}}{{1 - \sin x}} - \dfrac{1}{{\cos x}} = \tan x ?

Explanation

Solution

By using the basic trigonometric identity given below we can simplify the above expression that is cos⁡x1−sin⁡x−1cos⁡x=tan⁡x\dfrac{{\cos x}}{{1 - \sin x}} - \dfrac{1}{{\cos x}} = \tan x . In order to solve and simplify the given expression we have to use the above identity and express our given expression in the simplest form and thereby solve it.

Complete step by step solution:
To prove :-
cos⁡x1−sin⁡x−1cos⁡x=tan⁡x................(1)\dfrac{{\cos x}}{{1 - \sin x}} - \dfrac{1}{{\cos x}} = \tan x................(1)
Proof:-
Taking left hand side of the equation (1)(1) , we will get ,
cos⁡x1−sin⁡x−1cos⁡x\dfrac{{\cos x}}{{1 - \sin x}} - \dfrac{1}{{\cos x}}
Now , take L.C.M of the above expression , we will get as follow ,
⇒cos⁡x(cos⁡x)−1(1−sin⁡x)1−sin⁡x(cos⁡x) ⇒cos⁡2x−1+sin⁡x(1−sin⁡x)(cos⁡x)  \Rightarrow \dfrac{{\cos x(\cos x) - 1(1 - \sin x)}}{{1 - \sin x(\cos x)}} \\\ \Rightarrow \dfrac{{{{\cos }^2}x - 1 + \sin x}}{{(1 - \sin x)(\cos x)}} \\\
Now applying trigonometric identity that is cos⁡2x−1=−sin⁡2x{\cos ^2}x - 1 = - {\sin ^2}x ,
We will get the following ,
⇒−sin⁡2x+sin⁡x(1−sin⁡x)(cos⁡x)\Rightarrow \dfrac{{ - {{\sin }^2}x + \sin x}}{{(1 - \sin x)(\cos x)}}
Taking sin⁡x\sin xcommon from both the terms present in the numerator.
We will get the following ,
⇒sin⁡x(−sin⁡x+1)(1−sin⁡x)(cos⁡x)\Rightarrow \dfrac{{\sin x( - \sin x + 1)}}{{(1 - \sin x)(\cos x)}}
We can also write it as ,
⇒sin⁡x(1−sin⁡x)(1−sin⁡x)(cos⁡x)\Rightarrow \dfrac{{\sin x(1 - \sin x)}}{{(1 - \sin x)(\cos x)}}
Now we have to simplify the given equation as shown below ,
⇒sin⁡xcos⁡x\Rightarrow \dfrac{{\sin x}}{{\cos x}}
We know that ⇒sin⁡xcos⁡x=tan⁡x \Rightarrow \dfrac{{\sin x}}{{\cos x}} = \tan x ,
Therefore, we get ,
⇒tan⁡x\Rightarrow \tan x
Which is equal to R.H.S ,
Since L.H.S=R.H.S
Hence proved.

Note: Some other equations needed for solving these types of problem are:
cos⁡2x−1=−sin⁡2x{\cos ^2}x - 1 = - {\sin ^2}x and
⇒sin⁡xcos⁡x=tan⁡x\Rightarrow \dfrac{{\sin x}}{{\cos x}} = \tan x
Range of cosine and sine: [−1,1]\left[ { - 1,1} \right] ,
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.