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

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

Question

(B) Given that,$	heta = \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18)$.Using the property $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$,we have:$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given that,$	heta = \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18)$.Using the property $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$,we have:$	heta = 	an^{-1}(\frac{1}{7}) + 	an^{-1}(\frac{1}{8}) + 	an^{-1}(\frac{1}{18})$.First,apply the formula $	an^{-1}(x) + 	an^{-1}(y) = 	an^{-1}(\frac{x+y}{1-xy})$ for the first two terms:$	an^{-1}(\frac{1}{7}) + 	an^{-1}(\frac{1}{8}) = 	an^{-1}(\frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7 	imes 8}}) = 	an^{-1}(\frac{\frac{15}{56}}{\frac{55}{56}}) = 	an^{-1}(\frac{15}{55}) = 	an^{-1}(\frac{3}{11})$.Now,add the third term:$	heta = 	an^{-1}(\frac{3}{11}) + 	an^{-1}(\frac{1}{18}) = 	an^{-1}(\frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11 	imes 18}}) = 	an^{-1}(\frac{\frac{54+11}{198}}{\frac{198-3}{198}}) = 	an^{-1}(\frac{65}{195}) = 	an^{-1}(\frac{1}{3})$.Thus,$	an 	heta = \frac{1}{3}$,which implies $\cot 	heta = 3$.

If $\theta = \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18)$,then $\cot \theta$ is equal to

Similar Questions

If $a=\sin ^{-1}(\sin (5))$ and $b=\cos ^{-1}(\cos (5))$,then $a^2+b^2$ is equal to

If ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi ,$ then $\frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}} = $

If $\tan ^{-1} a+\tan ^{-1} b+\tan ^{-1} c=\pi$,then which of the following is true?

If $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$, then $x$ is equal to

If $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$,then $4x^2 - 4xy \cos \alpha + y^2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module