# Find the value of x if $cosec^{-1}x=2\cot^{-1}7+\cot^{-1}\frac{3}{4}$

$\begin{array}{1 1} \frac{125}{117} \\ \frac{-125}{117} \\ \frac{-117 }{125} \\ \frac{117}{125} \end{array}$

Toolbox:
• $$cot^{-1}x=tan^{-1}\large\frac{1}{x}$$
• $$2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}$$
• $$tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}$$
• $$tan^{-1}x=cosec^{-1}\large\frac{\sqrt{1+x^2}}{x}$$
R.H.S.
By substituting $$x=7\:and \:x=\large\frac{4}{3}$$in the formula of $$cot^{-1}x$$, R.H.S. becomes
$$2tan^{-1}\large\frac{1}{7}+tan^{-1}\large\frac{3}{4}$$

By taking $$x=\large\frac{1}{7},\large\frac{2x}{1-x^2}=\large\frac{2.\large\frac{1}{7}}{1-\large\frac{1}{49}}=\large\frac{2}{7}.\large\frac{49}{48}=\large\frac{7}{24}$$
$$\Rightarrow\:2tan^{-1}\large\frac{1}{7}=tan^{-1}\large\frac{7}{24}$$
$$2tan^{-1}\large\frac{1}{7}+tan^{-1}\large\frac{3}{4} = tan^{-1}\large\frac{7}{24}+tan^{-1}\large\frac{4}{3}$$

By taking$$x=\large\frac{7}{24}\:and\:y=\large\frac{4}{7}$$,
$$\large\frac{x+y}{1-xy}=\large\frac{\frac{7}{24}+\large\frac{4}{3}}{1-\large\frac{28}{72}}=\large\frac{117}{72}.\large\frac{72}{44}=\large\frac{117}{44}$$

Substituting the values in R.H.S we get
$$tan^{-1}\large\frac{7}{24}+tan^{-1}\large\frac{4}{3} = tan^{-1}\large\frac{\large\frac{7}{24}+\large\frac{4}{3}}{1-\large\frac{28}{72}}=tan^{-1}\large\frac{117}{44}$$

By taking $$x=\large\frac{117}{44},$$
$$\large\frac{\sqrt{1+x^2}}{x}=\large\frac{\sqrt{1+\large\frac{13689}{1936}}}{\large\frac{117}{44}}=\sqrt{\large\frac{15624}{1936}}.\large\frac{44}{117}=\large\frac{125}{117}$$
$$\Rightarrow\: tan^{-1}\large\frac{117}{44}= cosec^{-1}\large\frac{125}{117}$$

By substituting the value in the above equation we get
$$cosec^{-1}x= cosec^{-1}\frac{125}{117}$$
$$\Rightarrow\: x=\frac{125}{117}$$

edited Mar 19, 2013