Question

How do you simplify $\dfrac{2-{{\sec }^{2}}x}{{{\sec }^{2}}x}$?

Answer 1

Hint: We first multiply ${{\cos }^{2}}x$ to both the numerator and the denominator of the fraction $\dfrac{2-{{\sec }^{2}}x}{{{\sec }^{2}}x}$. Then we complete the multiplication and replace the value of ${{\cos }^{2}}x\times {{\sec }^{2}}x$ with 1 as $\cos x\times \sec x=1$. Then we use the theorem of multiple angles $\cos 2x=2{{\cos }^{2}}x-1$ and find the solution of the equation $\dfrac{2-{{\sec }^{2}}x}{{{\sec }^{2}}x}$ as 1.

Complete step by step answer:
We have been given to find the solution of the trigonometric function $\dfrac{2-{{\sec }^{2}}x}{{{\sec }^{2}}x}$.
We multiply both the numerator and the denominator of the fraction with ${{\cos }^{2}}x$.
Therefore, $\dfrac{2-{{\sec }^{2}}x}{{{\sec }^{2}}x}=\dfrac{{{\cos }^{2}}x\left( 2-{{\sec }^{2}}x \right)}{{{\cos }^{2}}x\times {{\sec }^{2}}x}$.
We know that in trigonometric identity relation $\cos x$ is inverse of $\sec x$.
This means $\cos x=\dfrac{1}{\sec x}$ which gives $\cos x\times \sec x=1$.
For the exponent theorem we get that for two numbers $a$ and $b$ we have ${{a}^{2}}\times {{b}^{2}}={{\left( ab \right)}^{2}}$.
In the multiplication of the equation $\dfrac{{{\cos }^{2}}x\left( 2-{{\sec }^{2}}x \right)}{{{\cos }^{2}}x\times {{\sec }^{2}}x}$, we multiply ${{\cos }^{2}}x$ with 2 and get $2{{\cos }^{2}}x$.
Then we multiply ${{\cos }^{2}}x$ with ${{\sec }^{2}}x$ and get ${{\cos }^{2}}x\times {{\sec }^{2}}x$. Same thing happens for denominator and we get ${{\cos }^{2}}x\times {{\sec }^{2}}x$.
Therefore, in the expression $\dfrac{{{\cos }^{2}}x\left( 2-{{\sec }^{2}}x \right)}{{{\cos }^{2}}x\times {{\sec }^{2}}x}=\dfrac{2{{\cos }^{2}}x-{{\cos }^{2}}x\times {{\sec }^{2}}x}{{{\cos }^{2}}x\times {{\sec }^{2}}x}$, the value of ${{\cos }^{2}}x\times {{\sec }^{2}}x$ can be replaced as 1.
The value of ${{\cos }^{2}}x\times {{\sec }^{2}}x$ can be written as ${{\cos }^{2}}x\times {{\sec }^{2}}x={{\left( \cos x\times \sec x \right)}^{2}}$. We know that $\cos x\times \sec x=1$ which gives ${{\cos }^{2}}x\times {{\sec }^{2}}x={{\left( \cos x\times \sec x \right)}^{2}}=1$.
The simplified form of $\dfrac{2{{\cos }^{2}}x-{{\cos }^{2}}x\times {{\sec }^{2}}x}{{{\cos }^{2}}x\times {{\sec }^{2}}x}$ is
$\dfrac{2{{\cos }^{2}}x-{{\cos }^{2}}x\times {{\sec }^{2}}x}{{{\cos }^{2}}x\times {{\sec }^{2}}x}=\dfrac{2{{\cos }^{2}}x-1}{1}=2{{\cos }^{2}}x-1$.
Now we apply the theorem of multiple angles which gives us $\cos 2x=2{{\cos }^{2}}x-1$.
We can replace the value in the equation and get $\dfrac{2-{{\sec }^{2}}x}{{{\sec }^{2}}x}=\cos 2x$.
Therefore, the simplified form of $\dfrac{2-{{\sec }^{2}}x}{{{\sec }^{2}}x}$ is $\cos 2x$.

Note: The denominator value ${{\sec }^{2}}x$ in the function of $\dfrac{2-{{\sec }^{2}}x}{{{\sec }^{2}}x}$ can be converted to ${{\cos }^{2}}x$ through inverse theorem and then we can multiply it with $2-{{\sec }^{2}}x$ to solve the equation. The other form of the multiple angle theorem $\cos 2x$ is $\cos 2x=1-2{{\sin }^{2}}x$.