Question

How do you prove $\dfrac{{1 + \sin x}}{{1 - \sin x}} = {(\sec x + \tan x)^2}$ ?

Answer 1

Hint : In order to proof the above statement ,take LHS first and Multiply divide with $1 + \sin x$ and then use ${A^2} + {B^2} + 2AB = {\left( {A + B} \right)^2}$ in the numerator to expand and $(A + b)(A - B) = {A^2} - {B^2}$ in the denominator to proof that LHS =RHS

Complete step-by-step answer :
To prove: $\dfrac{{1 + \sin x}}{{1 - \sin x}} = {(\sec x + \tan x)^2}$
Proof:
Taking Left-hand Side,
$\Rightarrow \dfrac{{1 + \sin x}}{{1 - \sin x}}$
Multiply and divide with $1 + \sin x$ in above expression
$\Rightarrow \left( {\dfrac{{1 + \sin x}}{{1 - \sin x}}} \right)\left( {\dfrac{{1 + \sin x}}{{1 + \sin x}}} \right) \\\ \Rightarrow \dfrac{{{{\left( {1 + \sin x} \right)}^2}}}{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}} \\\$
In denominator using formula $(A + b)(A - B) = {A^2} - {B^2}$
And expanding numerator with the formula $$ ${A^2} + {B^2} + 2AB = {\left( {A + B} \right)^2}$
$\Rightarrow \dfrac{{1 + {{\sin }^2}x + 2\sin x}}{{1 - {{\sin }^2}x}}$
Using trigonometric identity ${\cos ^2}x = 1 - {\sin ^2}x$

$$\Rightarrow \dfrac{{1 + {{\sin }^2}x + 2\sin x}}{{{{\cos }^2}x}} \\\ \Rightarrow \dfrac{1}{{{{\cos }^2}x}} + \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{2\sin x}}{{{{\cos }^2}x}} \\\ \Rightarrow \dfrac{1}{{{{\cos }^2}x}} + \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{2\sin x}}{{\cos x}}\dfrac{1}{{\cos x}} \;$$

Using rule of trigonometry $\dfrac{1}{{{{\cos }^2}x}} = {\sec ^2}x\,or\,\dfrac{1}{{\cos x}} = \sec x$ , $\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = {\tan ^2}x$ or $\dfrac{{\sin x}}{{\cos x}} = \tan x$
$\Rightarrow {\sec ^{2x}} + {\tan ^2}x + 2\tan x\sec x$
Expression appear to be of the ${A^2} + {B^2} + 2AB = {\left( {A + B} \right)^2}$
$\Rightarrow {\left( {\sec x + \tan x} \right)^2}$
Taking Right-Hand Side
${(\sec x + \tan x)^2}$
Therefore, LHS=RHS
Hence Proved.

Note : 1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
2. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta$ for all values of $\theta$ and n $\in$ N.
3. Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$ for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$ for all x in its domain.
We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta$
Therefore, $\sin \theta$ and $\tan \theta$ and their reciprocals, $\cos ec\theta$ and $\cot \theta$ are odd functions whereas $\cos \theta$ and its reciprocal $\sec \theta$ are even functions.