Question

How do you prove \[\dfrac{1}{{{{\sec }^2}x}} + \dfrac{1}{{\cos e{c^2}x}} = 1\]?

Answer 1

Hint: In this question the trigonometric expression is given, for simplify this expression we will use the basic trigonometric identity. And there are some trigonometric identities as given below.
\[ \Rightarrow \dfrac{{\sin x}}{{\cos x}} = \tan x\]
\[ \Rightarrow \sin x = \dfrac{1}{{\csc x}}\]
\[ \Rightarrow \cos x = \dfrac{1}{{\sec x}}\]
\[ \Rightarrow \tan x = \dfrac{1}{{\cot x}}\]

Complete step by step answer:
In this question, the word trigonometric identity is used. First we know about trigonometric identities.
The trigonometric identities are defined as the equations that are true for right angle triangles.
We determine the trigonometric identities by using the right-angle triangle. Its angle is \[x\]. Here $b$ is the length of the base, $p$ is the height and $h$ is the length the hypotenuse.
There are three main functions in trigonometry: sin, cos and tan.
Then we write these functions according to the right angle triangle.
\[
  \sin x = \dfrac{p}{h} \\
  \cos x = \dfrac{b}{h} \\
  \tan x = \dfrac{p}{b} \\
 \]
Then,
\[\tan x\] is written as below.
\[ \Rightarrow \dfrac{{\sin x}}{{\cos x}} = \dfrac{{p/h}}{{b/h}} = \dfrac{p}{b} = \tan x\]
Then,
\[ \Rightarrow \tan x = \dfrac{{\sin x}}{{\cos x}}\]
And another trigonometric function is written as below (by using right angle triangle).
\[
  \csc x = \dfrac{h}{p} = \dfrac{1}{{\sin x}} \\
  \sec x = \dfrac{h}{b} = \dfrac{1}{{\cos x}} \\
  \cot x = \dfrac{b}{p} = \dfrac{1}{{\tan x}} \\
 \]

We can also write it as,
\[
  \sin x = \dfrac{p}{h} = \dfrac{1}{{\cos ecx}} \\
  \cos x = \dfrac{b}{h} = \dfrac{1}{{\sec x}} \\
  \tan x = \dfrac{p}{b} = \dfrac{1}{{\cot x}} \\
 \]

Now come to the question. The trigonometric form is given which is proving as below.
\[ \Rightarrow \dfrac{1}{{{{\sec }^2}x}} + \dfrac{1}{{\cos e{c^2}x}} = 1\]
We know that according to the right angle triangle \[\sec x = \dfrac{h}{b}\] and \[\cos ecx = \dfrac{h}{p}\] .
We take the left-hand side. And Put this value on the left hand side.
Then, the above equation is written as.
\[ \Rightarrow \dfrac{1}{{{{\left( {\dfrac{h}{b}} \right)}^2}}} + \dfrac{1}{{{{\left( {\dfrac{h}{p}} \right)}^2}}}\]
Now, we simplify the left-hand side.
Then,
\[ \Rightarrow \dfrac{1}{{\left( {\dfrac{{{h^2}}}{{{b^2}}}} \right)}} + \dfrac{1}{{\left( {\dfrac{{{h^2}}}{{{p^2}}}} \right)}}\]
\[ \Rightarrow \dfrac{{{b^2}}}{{{h^2}}} + \dfrac{{{p^2}}}{{{h^2}}}\]
We know that, according to Pythagoras theorem.
\[
   \Rightarrow \dfrac{b}{h} = \cos x \\
   \Rightarrow \dfrac{p}{h} = \sin x \\
 \]
Now, we put the value of \[\dfrac{b}{h} = \cos x\;\] and \[\dfrac{p}{h} = \sin x\] in the above equation.
Then,
\[ \Rightarrow {\cos ^2}x + {\sin ^2}x\]
It is also written as.
\[ \Rightarrow {\sin ^2}x + {\cos ^2}x\]
We know that \[{\sin ^2}x + {\cos ^2}x = 1\] Put that value in above.
Then,
Left hand side is.
\[ = 1\]
Left hand side \[ = \]right hand side
\[\therefore 1 = 1\]

Therefore, the left-hand side is equal to the right hand side. Then it is proved that
\[\dfrac{1}{{{{\sec }^2}x}} + \dfrac{1}{{\cos e{c^2}x}} = 1\]

Note:
If you want to find the result of trigonometric identities, then you are suggested to use the right-angle triangle and Pythagoras theorem. By using these two, we can easily find the result of any trigonometric identity.