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: Subtract 1 from both sides of the equation and use the trigonometric identity: \[{{\sec }^{2}}x-1={{\tan }^{2}}x\] to get a simplified form. Now, take the term containing tangent function on the L.H.S and the term containing secant function to the R.H.S. Square both the sides of the equality and convert the secant function into the tangent function once again by using the relation: \[{{\sec }^{2}}x=1+{{\tan }^{2}}x\]. Again square both sides of the equality and simplify the terms by taking it to the L.H.S, to get the answer.

Complete step-by-step answer:
In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles. They are distinct from triangle identities, which are identities involving angles and side lengths of a triangle.
Now let us come to the question. We have been given, $\sec x+{{\sec }^{2}}x=1$.
Therefore, on subtracting 1 from both sides of the equation, we get,
$\sec x+{{\sec }^{2}}x-1=1-1$
Using the trigonometric identity: \[{{\sec }^{2}}x-1={{\tan }^{2}}x\], we get,
$\sec x+{{\tan }^{2}}x=0$
Taking secant function to the R.H.S, we get,
${{\tan }^{2}}x=-\sec x$
On squaring both sides, we get,
$\begin{align}
  & {{\left( {{\tan }^{2}}x \right)}^{2}}={{\left( -\sec x \right)}^{2}} \\
 & \Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x \\
\end{align}$
Again using the identity: \[{{\sec }^{2}}x=1+{{\tan }^{2}}x\], we get,
${{\tan }^{4}}x=1+{{\tan }^{2}}x$
Again squaring both sides of the equality, we get,
$\begin{align}
  & {{\left( {{\tan }^{4}}x \right)}^{2}}={{\left( 1+{{\tan }^{2}}x \right)}^{2}} \\
 & \Rightarrow {{\tan }^{8}}x=1+{{\tan }^{4}}x+2{{\tan }^{2}}x \\
 & \Rightarrow {{\tan }^{8}}x+{{\tan }^{4}}x+2{{\tan }^{2}}x=1 \\
\end{align}$
Now, adding 1 on both sides of the equation, we get,
\[\begin{align}
  & {{\tan }^{8}}x+{{\tan }^{4}}x+2{{\tan }^{2}}x+1=1+1 \\
 & \Rightarrow {{\tan }^{8}}x+{{\tan }^{4}}x+2{{\tan }^{2}}x+1=2 \\
\end{align}\]
Hence, option (c) is the correct answer.

Note: One may note that, never try to simplify the given expression by changing $\sec \theta $ into $\dfrac{1}{\cos \theta }$ and $\tan \theta $ into $\dfrac{\sin \theta }{\cos \theta }$ in the whole expression. It will be really difficult for us to solve the question if we will apply the above conversion. You may note that we can use the quadratic equation to solve the above question by solving the quadratic equation: $\sec x+{{\sec }^{2}}x=1$, involving $\sec x$. From there we will get a value of $\sec x$ and we have to find the value of $\tan x$. Then we have to substitute this value of $\tan x$ in the expression ${{\tan }^{8}}x-{{\tan }^{4}}x-2{{\tan }^{2}}x+1$, to get the answer. But this process will involve some difficult calculation. So, you must use the process we have used in the solution.