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Question: How do you evaluate \[2\left( {{\tan }^{-1}}\left( \dfrac{1}{3} \right) \right)-{{\tan }^{-1}}\left(...

How do you evaluate 2(tan⁡−1(13))−tan⁡−1(−17)2\left( {{\tan }^{-1}}\left( \dfrac{1}{3} \right) \right)-{{\tan }^{-1}}\left( -\dfrac{1}{7} \right)?

Explanation

Solution

This question belongs to the topic of inverse trigonometric functions. For solving this type of question, we should know the formulas of the topic inverse trigonometric functions. We are going to use some formulas like:
tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right);
tan⁡−1(−x)=−(tan⁡−1x){{\tan }^{-1}}\left( -x \right)=-\left( {{\tan }^{-1}}x \right)
We are going to use the principle value of the inverse of tan for solving this question.

Complete step-by-step answer:
Let us solve this question.
In this question, we have to evaluate the term 2(tan⁡−1(13))−tan⁡−1(−17)2\left( {{\tan }^{-1}}\left( \dfrac{1}{3} \right) \right)-{{\tan }^{-1}}\left( -\dfrac{1}{7} \right) or we have to simply the given term.
The term that we have to simplify is
2(tan⁡−1(13))−tan⁡−1(−17)2\left( {{\tan }^{-1}}\left( \dfrac{1}{3} \right) \right)-{{\tan }^{-1}}\left( -\dfrac{1}{7} \right)
The above term can also be written as
\Rightarrow $$$${{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{3} \right)-{{\tan }^{-1}}\left( -\dfrac{1}{7} \right)
In the above equation, we are going to use the formula: tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy);x,y>0,xy<1{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right);x,y>0,xy<1
Using the above formula, we can write for the first two terms
\Rightarrow $$$${{\tan }^{-1}}\left( \dfrac{\dfrac{1}{3}+\dfrac{1}{3}}{1-\dfrac{1}{3}\times \dfrac{1}{3}} \right)-{{\tan }^{-1}}\left( -\dfrac{1}{7} \right)
The above term can also be written as
\Rightarrow $$$${{\tan }^{-1}}\left( \dfrac{\dfrac{2}{3}}{1-\dfrac{1}{9}} \right)-{{\tan }^{-1}}\left( -\dfrac{1}{7} \right)
(Here, we have written 13+13\dfrac{1}{3}+\dfrac{1}{3} as equal to23\dfrac{2}{3} and 13×13\dfrac{1}{3}\times \dfrac{1}{3} as equal to 19\dfrac{1}{9}. )
We can write the above equation as
\Rightarrow $$$${{\tan }^{-1}}\left( \dfrac{\dfrac{2}{3}}{\dfrac{8}{9}} \right)-{{\tan }^{-1}}\left( -\dfrac{1}{7} \right)
(Here, we have written 1−191-\dfrac{1}{9} after solving 1−19=9−19=891-\dfrac{1}{9}=\dfrac{9-1}{9}=\dfrac{8}{9} as 89\dfrac{8}{9})
The above term can also be written as
\Rightarrow $$$${{\tan }^{-1}}\left( \dfrac{2}{3}\times \dfrac{9}{8} \right)-{{\tan }^{-1}}\left( -\dfrac{1}{7} \right)
We can write the above equation as
\Rightarrow $$$${{\tan }^{-1}}\left( \dfrac{3}{4} \right)-{{\tan }^{-1}}\left( -\dfrac{1}{7} \right)
In the above equation, we will use the formula: tan⁡−1(−x)=−(tan⁡−1x),x∈R{{\tan }^{-1}}\left( -x \right)=-\left( {{\tan }^{-1}}x \right),x\in \mathbb{R}
Using this formula in the above term, we can write
\Rightarrow $$$${{\tan }^{-1}}\left( \dfrac{3}{4} \right)-\left( -{{\tan }^{-1}}\left( \dfrac{1}{7} \right) \right)
The above equation can also be written as
\Rightarrow $$$${{\tan }^{-1}}\left( \dfrac{3}{4} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)
Using the formula tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy);x,y>0,xy<1{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right);x,y>0,xy<1 in the above equation, we get
\Rightarrow $$$${{\tan }^{-1}}\left( \dfrac{\dfrac{3}{4}+\dfrac{1}{7}}{1-\dfrac{3}{4}\times \dfrac{1}{7}} \right)
We can write the term 34+17\dfrac{3}{4}+\dfrac{1}{7} after adding as 3×7+1×44×7\dfrac{3\times 7+1\times 4}{4\times 7}, which is also can be written as 21+428\dfrac{21+4}{28}
And the term 1−34×171-\dfrac{3}{4}\times \dfrac{1}{7} also can be written as 1−328=28−3281-\dfrac{3}{28}=\dfrac{28-3}{28}, which is also can be written as 2528\dfrac{25}{28}
The term tan⁡−1(34+171−34×17){{\tan }^{-1}}\left( \dfrac{\dfrac{3}{4}+\dfrac{1}{7}}{1-\dfrac{3}{4}\times \dfrac{1}{7}} \right) also can be written as
⇒tan⁡−1(21+4282528)\Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{21+4}{28}}{\dfrac{25}{28}} \right)
The above term can also be written as
⇒tan⁡−1(25282528)\Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{25}{28}}{\dfrac{25}{28}} \right)
The above term can also be written as
⇒tan⁡−1(1)\Rightarrow {{\tan }^{-1}}\left( 1 \right)
And we know that the principal value of tan⁡−1(1)=π4{{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}
So, the simplified value of the term 2(tan⁡−1(13))−tan⁡−1(−17)2\left( {{\tan }^{-1}}\left( \dfrac{1}{3} \right) \right)-{{\tan }^{-1}}\left( -\dfrac{1}{7} \right) will be π4\dfrac{\pi }{4}

Note: We should have a better knowledge in the topic of inverse trigonometric functions. Don’t forget the formulas which are in the following:
tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy);x,y>0,xy<1{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right);x,y>0,xy<1
tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy);x,y>0,xy<1{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right);x,y>0,xy<1
And, also don’t forget the properties of tan⁡−1x{{\tan }^{-1}}x like:
tan⁡−1(−x)=−(tan⁡−1x),x∈R{{\tan }^{-1}}\left( -x \right)=-\left( {{\tan }^{-1}}x \right),x\in \mathbb{R}
for solving these types of questions easily.