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Question: The number of integral values of \(x\) satisfying the equation \({\tan ^{ - 1}}\left( {3x} \right) +...

The number of integral values of xx satisfying the equation tan⁡−1(3x)+tan⁡−1(5x)=tan⁡−1(7x)+tan⁡−1(2x){\tan ^{ - 1}}\left( {3x} \right) + {\tan ^{ - 1}}\left( {5x} \right) = {\tan ^{ - 1}}\left( {7x} \right) + {\tan ^{ - 1}}\left( {2x} \right) is

Explanation

Solution

Hint: Simplify the given equation using the formula, tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right).

Complete step-by-step answer:
Now, simplify the equation by taking tan of the resultant equation on both sides, We will get an equation in xx. Solve the equation to find the value of xx. In the final answer, take only integral values of xx.

We are given, tan⁡−1(3x)+tan⁡−1(5x)=tan⁡−1(7x)+tan⁡−1(2x){\tan ^{ - 1}}\left( {3x} \right) + {\tan ^{ - 1}}\left( {5x} \right) = {\tan ^{ - 1}}\left( {7x} \right) + {\tan ^{ - 1}}\left( {2x} \right).
First let us simplify the given equations using the formula, 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(3x)+tan⁡−1(5x)=tan⁡−1(7x)+tan⁡−1(2x) ⇒tan⁡−1(3x+5x1−(3x)(5x))=tan⁡−1(7x+2x1−(7x)(2x)) ⇒tan⁡−1(8x1−15x2)=tan⁡−1(9x1−14x2)  \Rightarrow {\tan ^{ - 1}}\left( {3x} \right) + {\tan ^{ - 1}}\left( {5x} \right) = {\tan ^{ - 1}}\left( {7x} \right) + {\tan ^{ - 1}}\left( {2x} \right) \\\ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{3x + 5x}}{{1 - \left( {3x} \right)\left( {5x} \right)}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{7x + 2x}}{{1 - \left( {7x} \right)\left( {2x} \right)}}} \right) \\\ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{8x}}{{1 - 15{x^2}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{9x}}{{1 - 14{x^2}}}} \right) \\\
On taking tan on both sides, we get
(8x1−15x2)=(9x1−14x2)\left( {\dfrac{{8x}}{{1 - 15{x^2}}}} \right) = \left( {\dfrac{{9x}}{{1 - 14{x^2}}}} \right)
Simplify the equation and solve for the value of xx.
⇒8x(1−14x2)=9x(1−15x2) ⇒8x−112x3=9x−135x3 ⇒23x3−x=0 ⇒x(23x2−1)=0 ⇒x=0,x=±123  \Rightarrow 8x\left( {1 - 14{x^2}} \right) = 9x\left( {1 - 15{x^2}} \right) \\\ \Rightarrow 8x - 112{x^3} = 9x - 135{x^3} \\\ \Rightarrow 23{x^3} - x = 0 \\\ \Rightarrow x\left( {23{x^2} - 1} \right) = 0 \\\ \Rightarrow x = 0,x = \pm \dfrac{1}{{\sqrt {23} }} \\\
The only integer value of xx is 0.
Hence, there is only one integral values of xxsatisfying the equation tan⁡−1(3x)+tan⁡−1(5x)=tan⁡−1(7x)+tan⁡−1(2x){\tan ^{ - 1}}\left( {3x} \right) + {\tan ^{ - 1}}\left( {5x} \right) = {\tan ^{ - 1}}\left( {7x} \right) + {\tan ^{ - 1}}\left( {2x} \right).

Note: It is important to simplify question using the formula, tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) to avoid difficult calculations. After simplifying, we will consider only integral values of xx.
In the final answer, we have to write the number of integral solutions and not the actual integral value of xx, therefore, the answer will be 1 and not 0.