જો {x^2} + {y^2} + {z^2} = {r^2} હોય,તો {tan | vedclass

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

Question

(B) આપેલ પદાવલિ: $S = {	an ^{ - 1}}\left( {\frac{{xy}}{{zr}}} ight) + {	an ^{ - 1}}\left( {\frac{{yz}}{{xr}}} ight) + {	an ^{ - 1}}\left( {\fra

Vedclass · Accepted Answer

$\frac{\pi }{2}$

Vedclass · Answer

(B) આપેલ પદાવલિ: $S = {	an ^{ - 1}}\left( {\frac{{xy}}{{zr}}} ight) + {	an ^{ - 1}}\left( {\frac{{yz}}{{xr}}} ight) + {	an ^{ - 1}}\left( {\frac{{zx}}{{yr}}} ight)$.નિત્યસમ ${	an ^{ - 1}}A + {	an ^{ - 1}}B + {	an ^{ - 1}}C = {	an ^{ - 1}}\left( {\frac{{A + B + C - ABC}}{{1 - (AB + BC + CA)}}} ight)$ નો ઉપયોગ કરતા:$A = \frac{{xy}}{{zr}}, B = \frac{{yz}}{{xr}}, C = \frac{{zx}}{{yr}}$.અહીં $AB + BC + CA = \frac{{xy^2z}}{{xr^2}} + \frac{{yz^2x}}{{yr^2}} + \frac{{zx^2y}}{{zr^2}} = \frac{{y^2 + z^2 + x^2}}{{r^2}} = \frac{{r^2}}{{r^2}} = 1$.છેદ $1 - (AB + BC + CA) = 1 - 1 = 0$ હોવાથી,${	an ^{ - 1}}$ નો આર્ગ્યુમેન્ટ $\infty$ તરફ જાય છે.તેથી,$S = {	an ^{ - 1}}(\infty) = \frac{\pi }{2}$.

જો ${x^2} + {y^2} + {z^2} = {r^2}$ હોય,તો ${\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{zx}}{{yr}}} \right) = $

Similar Questions

જો $y = \cot^{-1} \left( \frac{1 + \sin 5x}{\cos 5x} \right)$ હોય, તો $\frac{dy}{dx}$ નું મૂલ્ય છે

$\tan ^2(\sec ^{-1} 4) + \cot ^2(\operatorname{cosec}^{-1} 3)$ નું મૂલ્ય શોધો.

$\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right) = . . . . .$

$x$ ના મૂલ્યોનો ગણ શોધો જેથી $\tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right)$ થાય.

$\sin^{-1}(\sin \frac{7\pi}{6}) + \cos^{-1}(\cos \frac{7\pi}{6}) + \tan^{-1}(\tan \frac{2\pi}{3})$ ની કિંમત શોધો...

Vedclass Test Series

Exam Paper Generator

Online Exam Module