Question

If $\sec x + {\sec ^2}x = 1$ then the value of ${\tan ^8}x - {\tan ^4}x - 2{\tan ^2}x + 1$ will be equal to
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: First of all this is a very simple and a very easy problem. In order to solve this problem we need to have some basic knowledge of trigonometry, which includes basic trigonometric identities and basic trigonometric formulas. Along with this we also need to understand and should be able to solve simple mathematical equations.
Here the trigonometric identity which is used here is as given below:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1$
Hence by rearranging the terms, the above expression becomes, as given below:
$ \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta $

Complete step-by-step solution:
We should know that one of the trigonometric identities which is used here is given below:
$ \Rightarrow {\sec ^2}x - {\tan ^2}x = 1$
$ \Rightarrow {\sec ^2}x - 1 = {\tan ^2}x$
Hence ${\sec ^2}x - 1 = {\tan ^2}x$,
$\therefore {\tan ^2}x = {\sec ^2}x - 1$
Given that $\sec x + {\sec ^2}x = 1$, which can be transformed as below:
$ \Rightarrow \sec x + {\sec ^2}x = 1$
$ \Rightarrow \sec x = 1 - {\sec ^2}x$
The Right hand side of the above expression can be re-written as given below:
\[ \Rightarrow \sec x = - (1 - {\sec ^2}x)\]
\[ \Rightarrow \sec x = - {\tan ^2}x\]
Hence we obtained that \[\sec x = - {\tan ^2}x\], now squaring this equation on both sides as given below:
$ \Rightarrow {\left( {\sec x} \right)^2} = {\left( { - {{\tan }^2}x} \right)^2}$
$ \Rightarrow {\sec ^2}x = {\tan ^4}x$
Now the above equation can be written by substituting with one of the trigonometric identity which is :
$ \Rightarrow {\sec ^2}x - {\tan ^2}x = 1$
$\therefore {\sec ^2}x = 1 + {\tan ^2}x$
Now substituting the above expression in the equation ${\sec ^2}x = {\tan ^4}x$, replacing ${\sec ^2}x$ with $1 + {\tan ^2}x$, which is given below:
$ \Rightarrow {\sec ^2}x = {\tan ^4}x$
$ \Rightarrow 1 + {\tan ^2}x = {\tan ^4}x$
Now squaring the above equation on both sides as given below:
$ \Rightarrow {\left( {1 + {{\tan }^2}x} \right)^2} = {\left( {{{\tan }^4}x} \right)^2}$
$ \Rightarrow 1 + {\tan ^4}x + 2{\tan ^2}x = {\tan ^8}x$
Rearranging the above equation, as given below:
$ \Rightarrow 1 = {\tan ^8}x - {\tan ^4}x - 2{\tan ^2}x$
Now adding 1 on both sides, as given below:
$ \Rightarrow 1 + 1 = {\tan ^8}x - {\tan ^4}x - 2{\tan ^2}x + 1$
$ \Rightarrow 2 = {\tan ^8}x - {\tan ^4}x - 2{\tan ^2}x + 1$
Hence we obtained the value of ${\tan ^8}x - {\tan ^4}x - 2{\tan ^2}x + 1$ to be, as given below:
$ \Rightarrow {\tan ^8}x - {\tan ^4}x - 2{\tan ^2}x + 1 = 2$
The value of ${\tan ^8}x - {\tan ^4}x - 2{\tan ^2}x + 1$ is 2.

Option C is the correct answer.

Note: While solving this problem we should understand that we are substituting in this equation ${\sec ^2}x = {\tan ^4}x$, in place of ${\sec ^2}x$, replacing ${\sec ^2}x$ with $1 + {\tan ^2}x$. There is a chance that we might be able to confuse while substituting this. One should take care. We have to remember all the trigonometric identities such as: ${\sin ^2}x + {\cos ^2}x = 1,{\sec ^2}x - {\tan ^2}x = 1$ and $\cos e{c^2}x - {\cot ^2}x = 1$.