Question

# If 3 tan-1x + cot-1x = π, then x equals ​(a) 0 (b) 1 (c) -1 (d) $\frac{1}{2}$

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Solution

## $3{\mathrm{tan}}^{-1}x+{\mathrm{cot}}^{-1}x=\mathrm{\pi }\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}x+{\mathrm{cot}}^{-1}x=\mathrm{\pi }\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{tan}}^{-1}x+\frac{\mathrm{\pi }}{2}=\mathrm{\pi }\left[{\mathrm{tan}}^{-1}x+{\mathrm{cot}}^{-1}x=\frac{\mathrm{\pi }}{2}\right]$ $⇒2{\mathrm{tan}}^{-1}x=\mathrm{\pi }-\frac{\mathrm{\pi }}{2}=\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}x=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒x=\mathrm{tan}\frac{\mathrm{\pi }}{4}=1$ Thus, the value of x is 1. Hence, the correct answer is option (b).

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