Question

How do you simplify ${\cot ^2}x\left( {{{\tan }^2}x + 1} \right)$?

Answer 1

To simplify ${\cot ^2}x\left( {{{\tan }^2}x + 1} \right)$, we will first use the identity ${\sec ^2}x - {\tan ^2}x = 1$, to simplify the given expression to ${\cot ^2}x\left( {{{\sec }^2}x} \right)$. Then we will convert the given equation into $\sin x$ and $\cos x$. As we know that $\cos e{c^2}x = \dfrac{1}{{{{\sin }^2}x}}$. Using this we will further simplify it to find the result.

Complete step by step answer:
Given in the question, we have to simplify ${\cot ^2}x\left( {{{\tan }^2}x + 1} \right) - - - (1)$.
As we know from trigonometric identity that ${\sec ^2}x - {\tan ^2}x = 1$ i.e.,
$\Rightarrow 1 + {\tan ^2}x = {\sec ^2}x - - - (2)$
Putting $(2)$ in $(1)$, we get
$\Rightarrow {\cot ^2}x\left( {{{\tan }^2}x + 1} \right) = {\cot ^2}x\left( {{{\sec }^2}x} \right)$
Now to simplify we will convert the given equation into $\sin x$ and $\cos x$.
As we know, $\cot x = \dfrac{{\cos x}}{{\sin x}}$ and $\sec x = \dfrac{1}{{\cos x}}$.
Using this, we get
$\Rightarrow {\cot ^2}x\left( {{{\tan }^2}x + 1} \right) = {\left( {\dfrac{{\cos x}}{{\sin x}}} \right)^2} \times {\left( {\dfrac{1}{{\cos x}}} \right)^2}$
On simplification, we get
$\Rightarrow {\cot ^2}x\left( {{{\tan }^2}x + 1} \right) = \left( {\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right) \times \left( {\dfrac{1}{{{{\cos }^2}x}}} \right)$
Cancelling the common terms from the numerator and the denominator, we get
$\Rightarrow {\cot ^2}x\left( {{{\tan }^2}x + 1} \right) = \dfrac{1}{{{{\sin }^2}x}}$
As we know that $\cos e{c^2}x = \dfrac{1}{{{{\sin }^2}x}}$. Using this, we get
$\Rightarrow {\cot ^2}x\left( {{{\tan }^2}x + 1} \right) = \cos e{c^2}x$
Therefore, on simplification of ${\cot ^2}x\left( {{{\tan }^2}x + 1} \right)$, we get $\cos e{c^2}x$.

Note:

Trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two sides. Trigonometric formulas are considered only for right angled triangles. We have three sides namely hypotenuse, perpendicular and base in a right-angled triangle. Hypotenuse is the longest side, side opposite to the angle is perpendicular and base is the side where both hypotenuse and perpendicular rests. Basically, there are six ratios for finding the elements in trigonometry.
Here, we have used the trigonometric identity ${\sec ^2}x - {\tan ^2}x = 1$. Trigonometric identities are equalities that involve trigonometric functions. An identity is an equation which is always true, no matter what values are substituted whereas an equation may not be true for some values that are substituted. There are many other identities that we can use according to the question to simplify the expression.