Question

How do you verify \[\dfrac{\cos x}{1-\sin x}=\sec x+\tan x\] ?

Answer 1

Hint: Since this question comprises the trigonometric function, so we prove this problem by using various trigonometric identities and formulas. We start solving this problem by using LHS. Firstly, we rationalize the denominator and then use algebraic identity $(~a+b)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ and distributive property $a(b+c)=ab+ac$in the equation. Doing further calculations, we put trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ in the denominator and at last, we put trigonometric formula $\cos x=\dfrac{1}{\sec x}$, and$\tan x=\dfrac{\sin x}{\cos x}$, to get the required RHS.

Complete step by step answer:
According to the question, we have to prove that: \[\dfrac{\cos x}{1-\sin x}=\sec x+\tan x\] .
Let’s start this problem by solving LHS, that is
LHS: \[\dfrac{\cos x}{1-\sin x}\]
 On rationalizing the denominator, that is multiplying the numerator and denominator by $(1+\sin x)$, we get
\[\Rightarrow \left( \dfrac{\cos x}{1-\sin x} \right).\left( \dfrac{\text{1+}\sin x}{\text{1+sin }x} \right)\]
\[\Rightarrow \dfrac{\cos x(1+\sin x)}{(1-\sin x)(1+\sin x)}\]
By using algebraic identity $(~a+b)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$in the denominator and distributive property $a(b+c)=ab+ac$ in the numerator, we get
\[\Rightarrow \dfrac{\cos x+\cos x.\sin x}{1-{{\sin }^{2}}x}\]
Put trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ in the denominator of the above equation, we get
$\Rightarrow \dfrac{\cos x+\cos x.\sin x}{{{\cos }^{2}}x}$
Now, we split the denominator to the addition, that is,
$\begin{align}
  & \Rightarrow \dfrac{\cos x}{{{\cos }^{2}}x}+\dfrac{\cos x.\sin x}{{{\cos }^{2}}x} \\
 & \Rightarrow \dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x} \\
\end{align}$
Thus, using the trigonometric formula $\cos x=\dfrac{1}{\sec x}$ and $\tan x=\dfrac{\sin x}{\cos x}$ , we get
$\Rightarrow \sec x+\tan x=RHS$
Hence Proved.

Note:
We should perform each step carefully to avoid calculation mistakes. We should always remember that in rationalizing the denominator, we have to multiply and divide it by (1+sin x) instead of (1-sin x). One of the alternative methods for solving this problem is starting the solution using RHS. Firstly, we put the trigonometric formula $\sec x=\dfrac{1}{\cos x}$ and $\tan x=\dfrac{\sin x}{\cos x}$ in the equation, and take LCM of the same. After that, rationalize the denominator and use distributive identity $(a+b).c=ac+bc$ in the equation. After further calculations, put the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and use algebraic identity $(a-b)(a+b)={{a}^{2}}-{{b}^{2}}$to get the required statement that is LHS.
Let $RHS=\sec x+\tan x$
We put trigonometric identity $\sec x=\dfrac{1}{\cos x}$ and $\tan x=\dfrac{\sin x}{\cos x}$ in the above equation, we get
$\Rightarrow \dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}$
Take LCM of the above equation, we get
$\Rightarrow \dfrac{1+\sin x}{\cos x}$
Now, rationalize the denominator of the above equation, that is multiply and divide it by (cos x), we get
$\Rightarrow \dfrac{1+\sin x}{\cos x}.\left( \dfrac{\cos x}{\cos x} \right)$
$\Rightarrow \dfrac{(1+\sin x).\cos x}{\cos x.\cos x}$
Using distributive property $(a+b).c=ac+bc$ in the numerator, we get
$\Rightarrow \dfrac{\cos x+\sin x.\cos x}{{{\cos }^{2}}x}$
Put trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ in the denominator to get
$\Rightarrow \dfrac{\cos x+\cos x.\sin x}{1-{{\sin }^{2}}x}$
Now, using algebraic identity $(a-b)(a+b)={{a}^{2}}-{{b}^{2}}$in the denominator, we get
$\Rightarrow \dfrac{\cos x(1+\sin x)}{(1+\sin x)(1-\sin x)}$
$\Rightarrow \dfrac{\cos x}{1-\sin x}=LHS$
Hence Proved.