Question

Evaluate: $\sec (\cos e{{c}^{-1}}x)$ is equal to
1. $\cos ec({{\sec }^{-1}}x)$
2. $\cot x$
3. $\pi$
4. None of these

Answer 1

In order to solve this problem,Firstly we will take the value inside the bracket equal to some constant and try to get the value of the whole function in the unknown variable form. Then again we will do the same thing for the options given to us and check which among them it is equal to.

Complete answer: The trigonometric function is given as follows,
$\sec (\cos e{{c}^{-1}}x)$…..$\left( 1 \right)$
So we will start by finding the above function value in $x$ .
Let us take,
$\cos e{{c}^{-1}}x=A$….$\left( 2 \right)$
Putting above value in equation (1) we get,
$\sec A$
Now we will find the above value using equation (2)
Taking the inverse function on the right we get,
$\Rightarrow x=\cos ecA$
We know cosecant is inverse of sine function so we get,
$\Rightarrow x=\dfrac{1}{\sin A}$
$\Rightarrow \sin A=\dfrac{1}{x}$
Using the identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ we get,
$\Rightarrow \dfrac{1}{{{x}^{2}}}+{{\cos }^{2}}A=1$
$\Rightarrow {{\cos }^{2}}A=1-\dfrac{1}{{{x}^{2}}}$
Square root both side,
$\Rightarrow \cos A=\sqrt{1-\dfrac{1}{{{x}^{2}}}}$
$\Rightarrow \cos A=\sqrt{\dfrac{{{x}^{2}}-1}{{{x}^{2}}}}$
Now as we know secant is inverse of cosine so we get,
$\Rightarrow \sec A=\sqrt{\dfrac{{{x}^{2}}}{{{x}^{2}}-1}}$
That means
$\sec (\cos e{{c}^{-1}}x)=\sqrt{\dfrac{{{x}^{2}}}{{{x}^{2}}-1}}$….$\left( 3 \right)$
Next we will take option (1) value which is,
$\cos ec({{\sec }^{-1}}x)$…..$\left( 4 \right)$
Again take,
${{\sec }^{-1}}x=B$….$\left( 5 \right)$
Putting in equation (4) we get,
$\cos ecB$
Now we will find the value of above function by using equation (5)
Taking inverse value on right side we get,
$\Rightarrow x=\sec B$
As cosine is inverse of secant we get,
$\Rightarrow \cos B=\dfrac{1}{x}$
Using the identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ we get,
$\Rightarrow {{\sin }^{2}}B+\dfrac{1}{{{x}^{2}}}=1$
$\Rightarrow {{\sin }^{2}}B=1-\dfrac{1}{{{x}^{2}}}$
Square root both side,
$\Rightarrow \sin B=\sqrt{1-\dfrac{1}{{{x}^{2}}}}$
$\Rightarrow \sin B=\sqrt{\dfrac{{{x}^{2}}-1}{{{x}^{2}}}}$
Now as we know cosecant is inverse of sine so we get,
$\Rightarrow \cos ecB=\sqrt{\dfrac{{{x}^{2}}}{{{x}^{2}}-1}}$
That means,
$\cos ec({{\sec }^{-1}}x)=\sqrt{\dfrac{{{x}^{2}}}{{{x}^{2}}-1}}$…$\left( 6 \right)$
So from equation (3) and (6) we get,
$\sec (\cos e{{c}^{-1}}x)=\cos ec({{\sec }^{-1}}x)$
Hence the correct option is (1).

Note:
In this type of question we should always let the inverse part equal to some variable as it becomes easy to find the value of trigonometric functions in terms of any variables. We didn’t prove for other options as along the solution we can see that no other option can be equal to the function given.