Question

How do you find the exact value of \[\tan^{- 1}\left( \tan\left( \dfrac{3\pi}{4} \right) \right)\] ?

Answer 1

Hint: In this question, we need to find the value of \[\tan^{- 1}\left( \tan\left( \dfrac{3\pi}{4} \right) \right)\] . The basic trigonometric functions are sine , cosine and tangent . Tangent function is nothing but a ratio of the opposite side of a right angle to the adjacent of the right angle. Mathematically, Pi (\[\pi\]) is Greek letter and is a mathematical constant . In trigonometry, the value of \[\pi\] is \[180^{o}\] . Tangent inverse is nothing but it is an inverse of the Trigonometric function tangent . First, we can consider the given expression as \[y\] then, we need to substitute the value of π in the given expression ,thus by solving it we get the value of \[y\] . Then we need to check whether the value lies in the range of \[tan^{-1}\] .
Formula Used:
\[\tan (180^{o} - \theta)\ = \tan ( - \theta)\]
\[\tan ( - \theta)\ = - \tan (\theta)\]

Complete step-by-step solution:
Let us consider
\[y = \tan^{- 1}\left( \tan\left( \dfrac{3\pi}{4} \right) \right)\]
By taking the \[tan^{-1}\] on the left side to opposite side it becomes Tan,
\[tany = tan\left( \dfrac{3\pi}{4} \right)\]
By substituting the value of \[\pi\] ,
We get,
\[\Rightarrow \tan y = tan\left( \dfrac{3\left( 180^{o} \right)}{4} \right)\]
On simplifying,
We get,
\[\Rightarrow\tan y = tan(135^{o})\]
The range of \[\tan^{- 1}\] is \[\left( - \dfrac{\pi}{2},\dfrac{\pi}{2} \right)\]
The value of \[\dfrac{\pi}{2}\] is equal to \[90^{o}\] .
Hence \[y = 135^{o}\] is not possible. Thus we can find the value of \[tan(135^{o})\] with the help of other angles of tangent.
We can rewrite \[135^{o}\] as \[(180^{o} – 45^{o})\]
Thus ,
\[\Rightarrow \tan y = tan(180^{o} – 45^{o})\]
We know that \[tan(180^{o} - \theta)\ = - tan\ \theta\]
\[\Rightarrow \tan y = -tan( -45^{o})\]
Since tangent is an odd function \[tan( - \theta)\ = - tan(\theta)\]
\[\Rightarrow \tan y = - tan(45^{o})\]
We can also rewrite \[45^{o}\] as \[\dfrac{\pi}{4}\] ,
\[\tan y = tan( - 45^{o})\]
\[\Rightarrow \tan y = \tan\left( - \dfrac{\pi}{4} \right)\]
Hence \[y = - \dfrac{\pi}{4}\] which is in the range of \[\left( - \dfrac{\pi}{2},\dfrac{\pi}{2} \right)\]
Thus \[y = \tan^{- 1}\left( \dfrac{tan3\pi}{4} \right) = - dfrac{ \pi}4\]
Therefore the value of \[\tan^{- 1}\left( \dfrac{tan3\pi}{4} \right) = -\dfrac{ \pi}{4}\]
Final answer :
The value of \[\tan^{- 1}\left( \dfrac{\tan 3\pi}{4} \right) = - \dfrac{ \pi}4\]

Note: The concept used in this problem is trigonometric identities and ratios. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the use of trigonometric functions. For the tangent of \[45\] degrees , we use the abbreviation of the tangent as tan of the trigonometric function together with the degree symbol \[^{o}\], and write it as \[\tan (45^{o})\] . Geometrically, \[\tan (45^{o})\] lies in the first quadrant. Hence the value of \[\tan (45^{o})\] is non-negative.