Question

How do you simplify the expression $1 + {\tan ^2}x$?

Answer 1

In the given question, we have been asked to simplify an expression involving the sum of a constant with a trigonometric ratio, with the trigonometric ratio raised to the second power, i.e., with the trigonometric ratio being squared. We know that ${\sin ^2}x + {\cos ^2}x = 1$, and we are going to use this relation to transform it to the form we want.

Formula Used:
We are going to use the relation between the squares of sine and cosine, which is
${\sin ^2}x + {\cos ^2}x = 1$

Complete step-by-step answer:
We need to simplify the expression $1 + {\tan ^2}x$.
We remember the basic formula ${\sin ^2}x + {\cos ^2}x = 1$.
Dividing the both sides of the equation by ${\cos ^2}x$, we get,
$\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} = \dfrac{1}{{{{\cos }^2}x}}$
Now, we know that $\dfrac{{\sin x}}{{\cos x}} = \tan x \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = {\tan ^2}x$ and $\dfrac{1}{{\cos x}} = \sec x \Rightarrow \dfrac{1}{{{{\cos }^2}x}} = {\sec ^2}x$
Hence, ${\tan ^2}x + 1 = {\sec ^2}x$.

Additional Information:
If the question would have asked for $1 + {\cot ^2}x$, we would have divided the two sides of the equation by ${\sin ^2}x$. This would have given us ${\cot ^2}x$ because $\dfrac{{\cos x}}{{\sin x}} = \cot x$. And hence, their sum $\left( {1 + {{\cot }^2}x} \right)$ comes out to be ${{\mathop{\rm cosec}\nolimits} ^2}x$.

Note: The only thing we need to remember is the formula of the sum of squares of two trigonometric ratios - sine and cosine, which is one. Then we can transform them to find the relation for the same between secant and tangent and, cosecant and cotangent.