frac{tan ^{-1}(sqrt{3})-sec ^{-1}(-2)}{operat | vedclass

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Question

(B) આપણે પ્રતિ-ત્રિકોણમિતીય વિધેયોના ગુણધર્મો જાણીએ છીએ: $\sec^{-1}(-x) = \pi - \sec^{-1}(x)$,$\operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{

Vedclass · Accepted Answer

$\frac{-4}{5}$

Vedclass · Answer

(B) આપણે પ્રતિ-ત્રિકોણમિતીય વિધેયોના ગુણધર્મો જાણીએ છીએ: $\sec^{-1}(-x) = \pi - \sec^{-1}(x)$,$\operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1}(x)$,અને $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$.વધુમાં,$	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$.$\sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.$\operatorname{cosec}^{-1}(-\sqrt{2}) = -\operatorname{cosec}^{-1}(\sqrt{2}) = -\frac{\pi}{4}$.$\cos^{-1}\left(\frac{-1}{2}ight) = \pi - \cos^{-1}\left(\frac{1}{2}ight) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.આ કિંમતોને પદાવલિમાં મૂકતા:$\frac{\frac{\pi}{3} - \frac{2\pi}{3}}{-\frac{\pi}{4} + \frac{2\pi}{3}} = \frac{-\frac{\pi}{3}}{\frac{-3\pi + 8\pi}{12}} = \frac{-\frac{\pi}{3}}{\frac{5\pi}{12}} = -\frac{\pi}{3} 	imes \frac{12}{5\pi} = -\frac{4}{5}$.

$\frac{\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)}{\operatorname{cosec}^{-1}(-\sqrt{2})+\cos ^{-1}\left(\frac{-1}{2}\right)}$ ની કિંમત શોધો.

Similar Questions

કિંમત શોધો: $\cot ^{ - 1}\left(\frac{xy + 1}{x - y}\right) + \cot ^{ - 1}\left(\frac{yz + 1}{y - z}\right) + \cot ^{ - 1}\left(\frac{zx + 1}{z - x}\right)$

જો $\cos^{-1} p + \cos^{-1} q + \cos^{-1} r = \pi$ હોય,તો $p^2 + q^2 + r^2 + 2pqr = $

$2 \tan ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=$

$\lim _{x \rightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)}$ ની કિંમત શોધો.

જો $2 \sin ^{-1} x-3 \cos ^{-1} x=4, x \in[-1,1]$ હોય,તો $2 \sin ^{-1} x+3 \cos ^{-1} x$ ની કિંમત શોધો.

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