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Question: The value of the expression \({{\tan }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)=\) (A) \(\dfrac{5...

The value of the expression tan⁡−1(tan⁡3π4)={{\tan }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)=
(A) 5π4\dfrac{5\pi }{4}
(B) π4\dfrac{\pi }{4}
(C) −π4-\dfrac{\pi }{4}
(D) None of these

Explanation

Solution

We solve this question by first considering the given expression tan⁡−1(tan⁡3π4){{\tan }^{-1}}\left( \tan \dfrac{3\pi }{4} \right). Then we substitute the value of tan⁡3π4\tan \dfrac{3\pi }{4} in it. Then we consider the formula, tan⁡−1(−x)=−tan⁡−1(x){{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}\left( x \right). Then we use it to simplify the value of a given expression. Then we use the principle value of the value present inside the tan⁡−1{{\tan }^{-1}} function present in the obtained value. Then we use the formula, tan⁡−1(tan⁡x)=x   for  x∈(−π2,π2){{\tan }^{-1}}\left( \tan x \right)=x\ \ \ for\ \ x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right) and substitute this value in the above obtained value and find the required value.

Complete step-by-step solution:
The expression we are given is tan⁡−1(tan⁡3π4){{\tan }^{-1}}\left( \tan \dfrac{3\pi }{4} \right).
First, let us substitute the value of tan⁡3π4\tan \dfrac{3\pi }{4} in the given expression.
As we know that tan⁡3π4=−1\tan \dfrac{3\pi }{4}=-1, the expression we have is converted as,
⇒tan⁡−1(tan⁡3π4)=tan⁡−1(−1)\Rightarrow {{\tan }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)={{\tan }^{-1}}\left( -1 \right)
Now let us consider the formula,
tan⁡−1(−x)=−tan⁡−1(x){{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}\left( x \right) for all x∈Rx\in R
So, using it we can write the above equation as,
⇒tan⁡−1(tan⁡3π4)=−tan⁡−1(1)............(1)\Rightarrow {{\tan }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)=-{{\tan }^{-1}}\left( 1 \right)............\left( 1 \right)
Now let us first find the value of tan⁡−1(1){{\tan }^{-1}}\left( 1 \right).
We know that, tan⁡π4=1\tan \dfrac{\pi }{4}=1.
So, we can write tan⁡−1(1){{\tan }^{-1}}\left( 1 \right) as,
⇒tan⁡−1(1)=tan⁡−1(tan⁡π4)\Rightarrow {{\tan }^{-1}}\left( 1 \right)={{\tan }^{-1}}\left( \tan \dfrac{\pi }{4} \right)
Substituting this value in equation (1) we get,
⇒tan⁡−1(tan⁡3π4)=−tan⁡−1(tan⁡π4)\Rightarrow {{\tan }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)=-{{\tan }^{-1}}\left( \tan \dfrac{\pi }{4} \right)
Now let us consider the formula,
tan⁡−1(tan⁡x)=x   for  x∈(−π2,π2){{\tan }^{-1}}\left( \tan x \right)=x\ \ \ for\ \ x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)
As π4∈(−π2,π2)\dfrac{\pi }{4}\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right), we get value in the above equation as,
⇒tan⁡−1(tan⁡3π4)=−π4\Rightarrow {{\tan }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)=-\dfrac{\pi }{4}
So, we get the value of tan⁡−1(tan⁡3π4){{\tan }^{-1}}\left( \tan \dfrac{3\pi }{4} \right) as −π4-\dfrac{\pi }{4}.
Hence the answer is Option C.

Note: The common mistake one makes while solving this problem is that one might assume the value of given expression as θ\theta and solve it as,
⇒tan⁡−1(tan⁡3π4)=θ ⇒tan⁡θ=tan⁡3π4 ⇒θ=3π4 \begin{aligned} & \Rightarrow {{\tan }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)=\theta \\\ & \Rightarrow \tan \theta =\tan \dfrac{3\pi }{4} \\\ & \Rightarrow \theta =\dfrac{3\pi }{4} \\\ \end{aligned}
Then by checking with the options one might mark the answer as Option D.
But it is wrong as the range of tan⁡−1x{{\tan }^{-1}}x is (−π2,π2)\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right). So, we need to consider the principal value of θ\theta above while solving.