Question

xed{\sf Trigonometry}" align="absmiddle" class="latex-formula">
Solve for x

$\\ \rm\rightarrowtail tan^{-1}x+2cot^{-1}x=\dfrac{2\pi}{3}$


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Answer 1

Step-by-step explanation:

Since, tan

−1

θ+cot

−1

θ=

2

π

∴tan

−1

x+2cot

−1

x=

3

2π

⇒cot

−1

x+

2

π

=

3

2π

⇒cot

−1

x=

3

2π

−

2

π

=

6

π

⇒x=

3

Answer 2

Answer:

→x=√3

Step-by-step explanation:

$tan^{-1} x+2cot^{-1} x=\frac{2\pi }{3} \\\\tan^{-1} x+2(\frac{\pi }{2}-tan^{-1} x)= \frac{2\pi }{3} \\\\-tan^{-1} x=\frac{2\pi }{3}-\pi \\\\-tan^{-1} x=-\frac{\pi }{3} \\\\tan^{-1} x=\frac{\pi }{3} \\\\x=tan\frac{\pi }{3} \\\\x=\sqrt{3}$