જો 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) આપેલ છે: $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$ ગુણધર્મ $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$ ($y > 0$ માટે) નો ઉપયોગ કરતા:

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) આપેલ છે: $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$ ગુણધર્મ $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$ ($y > 0$ માટે) નો ઉપયોગ કરતા: $	an ^{-1}(\frac{1}{7})+	an ^{-1}(\frac{1}{8})+	an ^{-1}(\frac{1}{18})=	an ^{-1}(\frac{1}{x})$ પ્રથમ,$	an ^{-1}(\frac{1}{7}) + 	an ^{-1}(\frac{1}{8})$ માટે $	an ^{-1}(A) + 	an ^{-1}(B) = 	an ^{-1}(\frac{A+B}{1-AB})$ સૂત્ર વાપરતા: $	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})$ હવે,ત્રીજું પદ ઉમેરતા: $	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})})$ $= 	an ^{-1}(\frac{\frac{54+11}{198}}{\frac{198-3}{198}}) = 	an ^{-1}(\frac{65}{195}) = 	an ^{-1}(\frac{1}{3})$ આમ,$	an ^{-1}(\frac{1}{3}) = 	an ^{-1}(\frac{1}{x})$,જેનો અર્થ છે કે $x = 3$.

જો $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$ હોય,તો $x$ ની કિંમત શોધો.

Similar Questions

$(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8} \Rightarrow x=$

$\cosh \left(\sinh ^{-1}(\sqrt{8})+\cosh ^{-1} 5\right)=$

જો $y = \cot^{-1} \left( \frac{1 + x}{1 - x} \right)$ હોય,તો $\frac{dy}{dx} = $

$2 \tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=$

$\tan ^{-1} \frac{3}{4} + \tan ^{-1} \frac{3}{5} - \tan ^{-1} \frac{8}{19} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module