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.