Question

How do you prove: $1 + {\tan ^2}x = {\sec ^2}x$ ?

Answer 1

The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\sec x = \dfrac{1}{{\cos x}}$ . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem and proving the result given to us.

Complete step by step answer:
In the given problem, we have to prove a trigonometric identity that can be further used in many questions and problems as a direct result and has wide ranging applications. For proving the desired result, we need to first know the definitions of all the six trigonometric ratios.Now, we need to make the left and right sides of the equation equal.
L.H.S. $= 1 + {\tan ^2}x$
As we know that $\tan x = \left( {\dfrac{{{\text{Opposite Side}}}}{{{\text{Adjacent Side}}}}} \right)$. So, we get,
$1 + {\left( {\dfrac{{{\text{Opposite Side}}}}{{{\text{Adjacent Side}}}}} \right)^2}$
$\Rightarrow 1 + \dfrac{{{{\left( {{\text{Opposite Side}}} \right)}^2}}}{{{{\left( {{\text{Adjacent Side}}} \right)}^2}}}$
$\dfrac{{{{\left( {{\text{Adjacent Side}}} \right)}^2} + {{\left( {{\text{Opposite Side}}} \right)}^2}}}{{{{\left( {{\text{Adjacent Side}}} \right)}^2}}}$
Using Pythagoras Theorem, we know,
${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Opposite Side}}} \right)^2} + {\left( {{\text{Adjacent Side}}} \right)^2}$
$\Rightarrow\dfrac{{{{\left( {{\text{Hypotenuse}}} \right)}^2}}}{{{{\left( {{\text{Adjacent Side}}} \right)}^2}}}$
Now, we use the basic trigonometric formula $\sec x = \dfrac{{\left( {{\text{Hypotenuse}}} \right)}}{{\left( {{\text{Adjacent Side}}} \right)}}$, so we have,
${\sec ^2}x = R.H.S.$

As $L.H.S=R.H.S$, hence the given identity proved.

Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations.