Question

Prove that ${\sec ^2}x + \cos e{c^2}x = {\sec ^2}x\cos e{c^2}x$ .

Answer 1

Hint: Here we are given with the equation with the trigonometric components and we are asked to prove this equation. To prove any equation, we need to consider both the sides separately and simplify it if we get the same value for both sides then we can say that the given equation satisfies. Another way of solving this type of problem is taking one side and simplify it to get the other side. Here we have trigonometric components in the given equation so we will simplify it using the standard trigonometric identities and formulae.
Formula:
Some formulae to remember:
$\sec x = \dfrac{1}{{\cos x}}$
$\cos ecx = \dfrac{1}{{\sin x}}$
$\sin 2x = 2\sin x\cos x$
${\sin ^2}x + {\cos ^2}x = 1$

Complete answer:
Given equation ${\sec ^2}x + \cos e{c^2}x = {\sec ^2}x\cos e{c^2}x$ .
We need to prove left- hand side equal to right- hand side.
First, we’ll consider left- hand side, i.e., ${\sec ^2}x + \cos e{c^2}x$
Putting $\sec x = \dfrac{1}{{\cos x}}$ and $\cos ecx = \dfrac{1}{{\sin x}}$ , we get,
${\sec ^2}x + \cos e{c^2}x = \dfrac{1}{{{{\cos }^2}x}} + \dfrac{1}{{{{\sin }^2}x}}$
Taking least common multiple of the denominators, we get,
$\dfrac{1}{{{{\cos }^2}x}} + \dfrac{1}{{{{\sin }^2}x}} = \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\cos }^2}x{{\sin }^2}x}}$
Now, we know, ${\sin ^2}x + {\cos ^2}x = 1$
So, the equation becomes, $\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\cos }^2}x{{\sin }^2}x}} = \dfrac{1}{{{{\cos }^2}x{{\sin }^2}x}}$
Now, multiply and divide by 4 on the right- hand side, we get,
$\dfrac{4}{{4{{\cos }^2}x{{\sin }^2}x}}$ which can be written as $\dfrac{4}{{{{\left( {2\cos x\sin x} \right)}^2}}}$ .
Also, we know that, $\sin 2x = 2\sin x\cos x$ , so replacing it in above equation, we get,
$\dfrac{4}{{{{\left( {\sin 2x} \right)}^2}}}$
Now, let’s consider right- hand side, ${\sec ^2}x\cos e{c^2}x$ .
Again, replace $\sec x = \dfrac{1}{{\cos x}}$ and $\cos ecx = \dfrac{1}{{\sin x}}$ , we get,
\[{\sec ^2}x\cos e{c^2}x = \dfrac{1}{{{{\cos }^2}x{{\sin }^2}x}}\] .
Again, multiply and divide by 4 on the right- hand side, we get,
$\dfrac{4}{{4{{\cos }^2}x{{\sin }^2}x}}$ which can be written as $\dfrac{4}{{{{\left( {2\cos x\sin x} \right)}^2}}}$
Again, applying the identity $\sin 2x = 2\sin x\cos x$ , we get,
$\dfrac{4}{{{{\left( {\sin 2x} \right)}^2}}}$
Hence, left- hand side is equal to right- hand side, i.e., $\dfrac{4}{{{{\left( {\sin 2x} \right)}^2}}} = \dfrac{4}{{{{\left( {\sin 2x} \right)}^2}}}$
Thus, ${\sec ^2}x + \cos e{c^2}x = {\sec ^2}x\cos e{c^2}x$ .

Note:
Easier way to solve any trigonometric function is to convert it in the form of $\cos \theta $ and $\sin \theta $ .
In this question, while simplifying left- hand side, before multiplying and dividing by $4$ , we can directly write $\dfrac{1}{{{{\cos }^2}x{{\sin }^2}x}} = {\sec ^2}x\cos e{c^2}x$ , which is equal to the right- hand side.
In some questions, it is necessary to simplify both left- hand side and right- hand side separately, so as to prove them equal and not just one of the sides to make the question easier.