Question

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

Answer 1

Hint: 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.