Question

How do you use the fundamental identities to simplify $\cos t(1 + {\tan ^2}t)?$

Answer 1

To simplify the given trigonometric expression, you have to use some fundamental identities of trigonometry related to tangent, secant and cosecant.
The following fundamental trigonometric identities will be used in order to simplify the given trigonometric expression:
${\sec ^2}x - {\tan ^2}x = 1$
$\cos x = \dfrac{1}{{\sec x}}$

Complete step by step solution:
In order to simplify the trigonometric expression $\cos t(1 + {\tan ^2}t)$ we have to know about some trigonometric identities related to tangent, cosecant and secant represented as $\tan ,\;\cos \;{\text{and}}\;\sec$ respectively.
For tangent and secant relation we have a fundamental trigonometric identity which contain both of them and give us the relation between them, this trigonometric identity is as following
${\sec ^2}x - {\tan ^2}x = 1$
So putting ${\sec ^2}t - {\tan ^2}t$ in the place of $1$ in the given trigonometric expression, we will get
$= \cos t(1 + {\tan ^2}t) \\\ = \cos t({\sec ^2}t - {\tan ^2}t + {\tan ^2}t) \\\ = \cos t \times {\sec ^2}t \\\$
Now we know that cosecant and secant are the multiplicative inverse of each other, i.e. $\cos x = \dfrac{1}{{\sec x}}$
So putting $\cos x = \dfrac{1}{{\sec x}}$ in the above expression, we will get
$= \cos t \times {\sec ^2}t \\\ = \dfrac{1}{{\sec t}} \times {\sec ^2}t \\\ = \sec t \\\$
So finally we get the simplified form of $\cos t(1 + {\tan ^2}t) = \sec t$ by use of two fundamental trigonometric identities which are ${\sec ^2}x - {\tan ^2}x = 1\;{\text{and}}\;\cos x = \dfrac{1}{{\sec x}}$

Note: We can simplify this by one more method as following:
$= \cos t(1 + {\tan ^2}t)$
We will convert each term in the expression $\cos t(1 + {\tan ^2}t)$ into sine and cosine
We know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$
$= \cos t(1 + {\tan ^2}t) \\\ = \cos t\left( {1 + {{\left( {\dfrac{{\sin t}}{{\cos t}}} \right)}^2}} \right) \\\ = \cos t\left( {\dfrac{{{{\cos }^2}t + {{\sin }^2}t}}{{{{\cos }^2}t}}} \right) \\\$
From fundamental trigonometric identities, we know the relation between sine and cosine as
${\sin ^2}x + {\cos ^2}x = 1$
Now putting the value $1$ in the place of ${\cos ^2}t + {\sin ^2}t$ in the expression $\cos t\left( {\dfrac{{{{\cos }^2}t + {{\sin }^2}t}}{{{{\cos }^2}t}}} \right)$, we will get
$= \cos t\left( {\dfrac{{{{\cos }^2}t + {{\sin }^2}t}}{{{{\cos }^2}t}}} \right) \\\ = \cos t\left( {\dfrac{1}{{{{\cos }^2}t}}} \right) \\\ = \dfrac{1}{{\cos t}} \\\$
Now we know that one divided by cosine is equals to its multiplicative inverse secant, i.e. $\sec x = \dfrac{1}{{\cos x}}$
$= \dfrac{1}{{\cos t}} \\\ = \sec x \\\$
So once again we have simplified $\cos t(1 + {\tan ^2}t)$ with the help of the relation between sine and cosine. Sometimes, changing each term in the expression into sine and cosine helps simplify the expression in an easy way.