Question

If ${\sec }^{-1}\left(\sqrt{1+x^2}\right)+{cosec}^{-1}\left(\frac{\sqrt{1+y^2}}{y}\right)+{\cot }^{-1}\left(\frac{1}{z}\right)=3\pi$ then, $x+y+z=?$

• A. $xyz$
• B. $2xyz$
• C. $xy{z}^2$
• D. $x^{2}yz$

Answer 1

The correct option is A

$xyz$

Explanation for the correct option

Step 1: Information required for the inverse trigonometric identities for the solution

First, use the identities,

${\mathbf{sec}}^{\mathbf{-}\mathbf{1}}\left(\theta \right)\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\left(\frac{1}{\theta }\right)\mathbf{,}\mathbf{}{\mathbf{cosec}}^{\mathbf{-}\mathbf{1}}\left(\theta \right)\mathbf{=}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\left(\frac{1}{\theta }\right)\mathbf{,}\mathbf{}\mathbf{\text{and}}\mathbf{}{\mathbf{cot}}^{{}^{\mathbf{-}\mathbf{1}}}\left(\theta \right)\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{1}{\theta }\right)$

Then, the given equation will become

${\cos }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)+{\sin }^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)+{\tan }^{-1}\left(z\right)=3\pi$

Now, use the identities,

${\mathbf{cos}}^{-1}\left(\frac{A}{\sqrt{A^2+B^2}}\right)\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{B}{A}\right)\mathbf{}\mathbf{\text{and}}\mathbf{}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\left(\frac{A}{\sqrt{1+A^2}}\right)\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(A\right)$

From, this we can write the above equation as,

${\tan }^{-1}\left(\frac{x}{1}\right)+{\tan }^{-1}\left(y\right)+{\tan }^{-1}\left(z\right)=3\pi$

Step 2: Determination of the value of $\mathit{x}\mathbf{+}\mathit{y}\mathbf{+}\mathit{z}$

Finally, use the identity,

${\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(A\right)\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(B\right)\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(C\right)\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{A+B+C-ABC}{1-AB-BC-CA}\right)$

$\begin{array}{rcl}\therefore {\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(y\right)+{\tan }^{-1}\left(z\right)& =& 3\pi \\ & \Rightarrow & {\tan }^{-1}\frac{[x+y+z-xyz]}{[1-xy-yz-zx]}=3\pi \\ & \Rightarrow & \frac{[x+y+z-xyz]}{[1-xy-yz-zx]}=\tan 3\pi \\ & \Rightarrow & \frac{[x+y+z-xyz]}{[1-xy-yz-zx]}=0\left[\because \tan 3\pi =0\right]\\ & \Rightarrow & [x+y+z-xyz]=0\\ & \Rightarrow & x+y+z=xyz\end{array}$

Hence, the correct option is (A).