Question

How do you prove the identity \[{\sec ^2}x - {\tan ^2}x = 1\] .

Answer 1

Hint: Just use the basic formulas of trigonometry and trigonometric identities. Draw a triangle with perpendicular, base and hypotenuse and mark them then use the formulas of basic trigonometry that is $ \sin x = \dfrac{{perpendicular}}{{hypotenuse}} $ or $ \cos x = \dfrac{{base}}{{hypotenuse}} $ .Then put these values in the equation and get the results.

Complete step-by-step answer:
We are given with the identity that \[{\sec ^2}x - {\tan ^2}x = 1\]
Let’s start from the base:
In a $ \vartriangle ABC $ right angled at B inclined at angle $ x $ degree, mark the base as $ b $ , perpendicular as $ p $ and hypotenuse as $ h $ .

Since from the basic trigonometric formulas we know that:
  $ \sin x = \dfrac{{perpendicular}}{{hypotenuse}} = \dfrac{p}{h} $
Similarly, for others:
$
  \cos x = \dfrac{{base}}{{hypotenuse}} = \dfrac{b}{h} \\
  \tan x = \dfrac{{perpendicular}}{{base}} = \dfrac{p}{b} \\
  \cos ecx = \dfrac{1}{{\sin x}} = \dfrac{1}{{\dfrac{p}{h}}} = \dfrac{h}{p} \\
  secx = \dfrac{1}{{\cos x}} = \dfrac{1}{{\dfrac{b}{h}}} = \dfrac{h}{b} \\
  \cot x = \dfrac{1}{{\tan x}} = \dfrac{1}{{\dfrac{p}{b}}} = \dfrac{b}{p} \;
  $
We are given the Identity $ {\sec ^2}x - {\tan ^2}x = 1 $ , just put the values of \[\sec x\] and \[\tan x\] one by one on the left and right hand sides.
Let's start with the left hand side that is L.H.S:
L.H.S $ = {\sec ^2}x - {\tan ^2}x $
Put the values:
L.H.S
 $
   = {\sec ^2}x - {\tan ^2}x \\
   = {\left( {\dfrac{h}{b}} \right)^2} - {\left( {\dfrac{p}{b}} \right)^2} \\
  $
Further solve the above equation and we get:
$
   = {\sec ^2}x - {\tan ^2}x \\
   = {\left( {\dfrac{h}{b}} \right)^2} - {\left( {\dfrac{p}{b}} \right)^2} \\
   = \dfrac{{{h^2}}}{{{b^2}}} - \dfrac{{{p^2}}}{{{b^2}}} \;
  $
Taking base common:
 \[
   = \dfrac{{{h^2}}}{{{b^2}}} - \dfrac{{{p^2}}}{{{b^2}}} \\
   = \dfrac{{{h^2} - {p^2}}}{{{b^2}}} \;
 \] ………………….(i)
From Pythagora's theorem we know that Sum of square of perpendicular and square of base is equal to the square of hypotenuse. In Mathematical form it is written as:
 \[{p^2} + {b^2} = {h^2}\]
So, it can also be written as \[{h^2} - {p^2} = {b^2}\]
Put the value obtained above in equation (i) and we get:
 \[
   = \dfrac{{{h^2} - {p^2}}}{{{b^2}}} \\
   = \dfrac{{{b^2}}}{{{b^2}}} = 1 \;
 \]
The value obtained on the Left Hand Side or L.H.S is \[1\] .
Let's check for the Right Hand Side.
So, R.H.S is already given as 1.
This means that L.H.S \[ = \] R.H.S.
Therefore, the given identity $ {\sec ^2}x - {\tan ^2}x = 1 $ is proved.

Note: Use of basic formulas only so it's very important to remember the formulas of trigonometry and their identities.
Check for both the left and right hand side if values are given on both sides.
Pythagora's Rule is important for these type of questions.