If cot ^{-1}(7)+cot ^{-1}(8)+cot ^{-1}(18)=co | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given: $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$ Using the property $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$ for $y > 0$: $	an ^

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given: $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$ Using the property $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$ for $y > 0$: $	an ^{-1}(\frac{1}{7})+	an ^{-1}(\frac{1}{8})+	an ^{-1}(\frac{1}{18})=	an ^{-1}(\frac{1}{x})$ First,apply $	an ^{-1}(A) + 	an ^{-1}(B) = 	an ^{-1}(\frac{A+B}{1-AB})$ for $	an ^{-1}(\frac{1}{7}) + 	an ^{-1}(\frac{1}{8})$: $	an ^{-1}(\frac{\frac{1}{7}+\frac{1}{8}}{1-(\frac{1}{7})(\frac{1}{8})}) = 	an ^{-1}(\frac{\frac{15}{56}}{\frac{55}{56}}) = 	an ^{-1}(\frac{15}{55}) = 	an ^{-1}(\frac{3}{11})$ Now,add the third term: $	an ^{-1}(\frac{3}{11}) + 	an ^{-1}(\frac{1}{18}) = 	an ^{-1}(\frac{\frac{3}{11}+\frac{1}{18}}{1-(\frac{3}{11})(\frac{1}{18})})$ $= 	an ^{-1}(\frac{\frac{54+11}{198}}{\frac{198-3}{198}}) = 	an ^{-1}(\frac{65}{195}) = 	an ^{-1}(\frac{1}{3})$ Thus,$	an ^{-1}(\frac{1}{3}) = 	an ^{-1}(\frac{1}{x})$,which implies $x = 3$.

If $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$,then the value of $x$ is

Similar Questions

The value of $2 \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{7}$ is:

If $\sec^{-1} x = \csc^{-1} y$,then $\cos^{-1} \frac{1}{x} + \cos^{-1} \frac{1}{y} = $

If $\alpha = \cos^{-1}\left(\frac{3}{5}\right)$ and $\beta = \tan^{-1}\left(\frac{1}{3}\right)$,where $0 < \alpha, \beta < \frac{\pi}{2}$,then $\alpha - \beta$ is equal to

If $\cos^{-1} x + \cos^{-1} y = 2\pi$,then $\sin^{-1} x + \sin^{-1} y$ is equal to

The value of $\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module