
If \[x>0,y>0,z>0,xy+yz+zx<1\] and \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \], then \[x+y+z\] is equal to
1.\[0\]
2.\[xyz\]
3.\[3xyz\]
4.\[\sqrt{\left( xyz \right)}\]
Answer
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Hint: In order to solve the problem, we will be considering the given equation i.e. \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \], then we will be transposing the term \[{{\tan }^{-1}}z\] from LHS to the RHS and solve accordingly on both sides by applying the expansion rules of trigonometry. Upon doing so, we will be obtaining the value of \[x+y+z\].
Complete step-by-step solution:
Let us have a brief regarding the trigonometric functions. The counter-clockwise angle between the initial arm and the terminal arm of an angle in standard position is called the principal angle. Its value is between \[{{0}^{\circ }}\] and \[{{360}^{\circ }}\]. The relationship between the angles and sides of a triangle are given by the trigonometric functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. These are the basic main trigonometric functions used.
Now let us find the value of \[x+y+z\].
We are given that \[x>0,y>0,z>0,xy+yz+zx<1\] and also \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \].
Now let us consider \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \].
Let us transpose the term \[{{\tan }^{-1}}z\] from LHS to the RHS, we get
\[\begin{align}
& {{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \\
& \Rightarrow {{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi -{{\tan }^{-1}}z \\
\end{align}\]
Now, we know that \[{{\tan }^{-1}}x+{{\tan }^{-1}}y=\left( \dfrac{x+y}{1-xy} \right)\] and also \[\pi -{{\tan }^{-1}}z={{\tan }^{-1}}\left( -z \right)\]
Now let us substitute accordingly and upon solving them, we get
\[\begin{align}
& \Rightarrow {{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi -{{\tan }^{-1}}z \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( -z \right) \\
\end{align}\]
Now let’s cancel out the common term i.e. \[{{\tan }^{-1}}\] on both sides and solve it, we get
\[\begin{align}
& \Rightarrow x+y=-z\left( 1-xy \right) \\
& \Rightarrow x+y=-z+xyz \\
\end{align}\]
\[\therefore x+y+z=xyz\]
Hence option 2 is the correct answer.
Note: While solving such problems, we must expand the terms correctly as incorrectly expanding is the most commonly committed error. Also, we must analyse before we proceed by considering which terms should be transposed and which are not supposed to because transposing the wrong terms can make the simplification complicated.
Complete step-by-step solution:
Let us have a brief regarding the trigonometric functions. The counter-clockwise angle between the initial arm and the terminal arm of an angle in standard position is called the principal angle. Its value is between \[{{0}^{\circ }}\] and \[{{360}^{\circ }}\]. The relationship between the angles and sides of a triangle are given by the trigonometric functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. These are the basic main trigonometric functions used.
Now let us find the value of \[x+y+z\].
We are given that \[x>0,y>0,z>0,xy+yz+zx<1\] and also \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \].
Now let us consider \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \].
Let us transpose the term \[{{\tan }^{-1}}z\] from LHS to the RHS, we get
\[\begin{align}
& {{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \\
& \Rightarrow {{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi -{{\tan }^{-1}}z \\
\end{align}\]
Now, we know that \[{{\tan }^{-1}}x+{{\tan }^{-1}}y=\left( \dfrac{x+y}{1-xy} \right)\] and also \[\pi -{{\tan }^{-1}}z={{\tan }^{-1}}\left( -z \right)\]
Now let us substitute accordingly and upon solving them, we get
\[\begin{align}
& \Rightarrow {{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi -{{\tan }^{-1}}z \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( -z \right) \\
\end{align}\]
Now let’s cancel out the common term i.e. \[{{\tan }^{-1}}\] on both sides and solve it, we get
\[\begin{align}
& \Rightarrow x+y=-z\left( 1-xy \right) \\
& \Rightarrow x+y=-z+xyz \\
\end{align}\]
\[\therefore x+y+z=xyz\]
Hence option 2 is the correct answer.
Note: While solving such problems, we must expand the terms correctly as incorrectly expanding is the most commonly committed error. Also, we must analyse before we proceed by considering which terms should be transposed and which are not supposed to because transposing the wrong terms can make the simplification complicated.
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