Question

The number of value(s) of `x`, for which `tan^(-1)(1/x) = pi+tan^(-1)x, 0 lt x lt 1`

Answer 1

➡️Solution:

To find the number of value(s) of x,

${tan}^{ - 1} \bigg( \frac{1}{x} \bigg) = \pi + {tan}^{ - 1} x \\ \\ {tan}^{ - 1} \bigg( \frac{1}{x} \bigg) - {tan}^{ - 1} x = \pi$
As we know that

${tan}^{ - 1} A - {tan}^{ - 1} B = {tan}^{ - 1}\bigg( \frac{A - B}{1 + AB} \bigg)$
So,apply this formula
${tan}^{ - 1} \bigg( \frac{ \frac{1}{x} - x}{1 + \frac{1}{x} x} \bigg) = \pi \\ \\ {tan}^{ - 1} \bigg( \frac{ \frac{1 - {x}^{2} }{x} }{2} \bigg) = \pi \\ \\ {tan}^{ - 1} \bigg(\frac{1 - {x}^{2} }{2x} \bigg) = \pi \\ \\ taking \: tan \: both \: sides \\ \\ \frac{1 - {x}^{2} }{2x} = tan \: \pi \\ \\ \frac{1 - {x}^{2} }{2x} = 0 \\ \\ 1 - {x}^{2} = 0 \\ \\ {x}^{2} = 1 \\ \\ x = ± 1$

Thus,x can be +1 and -1.

Hope it helps you