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Question: How do you prove\(\left( {\sec x + \tan x} \right)\left( {\dfrac{{1 - \sin x}}{{\cos x}}} \right) = ...

How do you prove(sec⁡x+tan⁡x)(1−sin⁡xcos⁡x)=1\left( {\sec x + \tan x} \right)\left( {\dfrac{{1 - \sin x}}{{\cos x}}} \right) = 1?

Explanation

Solution

Use the given below formulas to solve the problem
sec⁡x=1cos⁡x tan⁡x=sin⁡xcos⁡x  \sec x = \dfrac{1}{{\cos x}} \\\ \tan x = \dfrac{{\sin x}}{{\cos x}} \\\
The very first step to solve this problem, we will replace all the terms on the LHS with the required simple terms. After doing this, start with LHS and multiply and add the terms just like the operations mentioned. After that, replace 1−sin⁡2x1 - {\sin ^2}xby cos⁡2x{\cos ^2}xto get a fraction with both numerator and denominator as cos⁡2x{\cos ^2}x. Cancelling out those two we will get 1 as our answer which will be our RHS. Hence proved.

Complete step-by-step solution:
The given question we have is to prove(sec⁡x+tan⁡x)(1−sin⁡xcos⁡x)=1\left( {\sec x + \tan x} \right)\left( {\dfrac{{1 - \sin x}}{{\cos x}}} \right) = 1
Now, before starting off with the solution, we need to look for some trigonometric identities. And some of which we will use in this question are:-
sec⁡x=1cos⁡x tan⁡x=sin⁡xcos⁡x  \sec x = \dfrac{1}{{\cos x}} \\\ \tan x = \dfrac{{\sin x}}{{\cos x}} \\\
Considering the LHS. When we replace the values of sec⁡x\sec xandtan⁡x\tan xin the given problem, we will get:-
⇒(1cos⁡x+sin⁡xcos⁡x)(1−sin⁡xcos⁡x)\Rightarrow \left( {\dfrac{1}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}}} \right)\left( {\dfrac{{1 - \sin x}}{{\cos x}}} \right)
Solving the first bracket
⇒(1+sin⁡xcos⁡x)(1−sin⁡xcos⁡x)\Rightarrow \left( {\dfrac{{1 + \sin x}}{{\cos x}}} \right)\left( {\dfrac{{1 - \sin x}}{{\cos x}}} \right)
Multiplying and opening the brackets our simplified version will be
⇒(1+sin⁡x)(1−sin⁡x)cos⁡2x\Rightarrow \dfrac{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}{{{{\cos }^2}x}}
To move further, we will use one identity which is quite simple and easy to remember. And that is
⇒1−sin⁡2xcos⁡2x=cos⁡2xcos⁡2x ⇒1−sin⁡2xcos⁡2x=11 ⇒1−sin⁡2xcos⁡2x=1  \Rightarrow \dfrac{{1 - {{\sin }^2}x}}{{{{\cos }^2}x}} = \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} \\\ \Rightarrow \dfrac{{1 - {{\sin }^2}x}}{{{{\cos }^2}x}} = \dfrac{1}{1} \\\ \Rightarrow \dfrac{{1 - {{\sin }^2}x}}{{{{\cos }^2}x}} = 1 \\\
Using the above identity on the question. We will get
(1+sin⁡x)(1−sin⁡x)cos⁡2x=12−sin⁡2xcos⁡2x\dfrac{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}{{{{\cos }^2}x}} = \dfrac{{{1^2} - {{\sin }^2}x}}{{{{\cos }^2}x}}
Now, we know that
sin⁡2x+cos⁡2x=1 ∴1−sin⁡2x=cos⁡2x  {\sin ^2}x + {\cos ^2}x = 1 \\\ \therefore 1 - {\sin ^2}x = {\cos ^2}x \\\
Using the above value of 1−sin⁡2x1 - {\sin ^2}xin our question. We will get:-
⇒1−sin⁡2xcos⁡2x=cos⁡2xcos⁡2x ⇒1−sin⁡2xcos⁡2x=11 ⇒1−sin⁡2xcos⁡2x=1 \Rightarrow \dfrac{{1 - {{\sin }^2}x}}{{{{\cos }^2}x}} = \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} \\\ \Rightarrow \dfrac{{1 - {{\sin }^2}x}}{{{{\cos }^2}x}} = \dfrac{1}{1} \\\ \Rightarrow \dfrac{{1 - {{\sin }^2}x}}{{{{\cos }^2}x}} = 1 \\\

Hence, when we solve the LHS of the equation, we end up with a value which was our RHS. This thing proves that RHS=LHS and hence the question given to us is finally proved.

Note: Identities of the trigonometry are one of the most important things when you are solving problems. From basic identities to ratios and formulas, everything is used to solve one problem. Therefore, having these on your fingertips is the first thing you must practice. In this way you will easily ace this chapter.