Question

\( \operatorname{5tan}^{-1}(x)+3 \cot ^{-1}(x)=\frac{7 \pi}{4} \)

Answer 1

5 tan^-1x + 3 cot^-1x = \(\frac{7\pi}4\)

⇒ 2 tan^-1x + 3(tan^-1x + cot^-1x) = \(\frac{7\pi}4\)

⇒ 2 tan^-1x + \(\frac{3\pi}2\) = \(\frac{7\pi}4\)

⇒ 2 tan^-1x = \(\frac{7\pi}4\) - \(\frac{3\pi}2\) = \(\frac{7\pi-5\pi}4=\frac{\pi}4\)

⇒ tan^-1x = \(\frac{\pi}8\)

⇒ x = tan\((\frac{\pi}8)\) ≅ 0.414