tan left(sin ^{-1}left(frac{3}{5}right)+tan ^ | vedclass

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Question

(B) આપણે જાણીએ છીએ કે $\sin ^{-1}\left(\frac{3}{5}ight) = 	an ^{-1}\left(\frac{3}{\sqrt{1^2 - (3/5)^2}}ight) = 	an ^{-1}\left(\frac{3/5}{4/5}i

Vedclass · Accepted Answer

$\frac{17}{6}$

Vedclass · Answer

(B) આપણે જાણીએ છીએ કે $\sin ^{-1}\left(\frac{3}{5}ight) = 	an ^{-1}\left(\frac{3}{\sqrt{1^2 - (3/5)^2}}ight) = 	an ^{-1}\left(\frac{3/5}{4/5}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$.સૂત્ર $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$ નો ઉપયોગ કરતા:$	an \left[	an ^{-1}\left(\frac{3}{4}ight)+	an ^{-1}\left(\frac{2}{3}ight)ight]$$= 	an \left[	an ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}ight)ight]$$= 	an \left[	an ^{-1}\left(\frac{\frac{9+8}{12}}{1-\frac{6}{12}}ight)ight]$$= 	an \left[	an ^{-1}\left(\frac{17/12}{6/12}ight)ight]$$= 	an \left[	an ^{-1}\left(\frac{17}{6}ight)ight]$$= \frac{17}{6}$

$\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ ની કિંમત શોધો.

Similar Questions

જો $0 < x < \frac{1}{2}$ માટે $y = 2\sin^{-1} \sqrt{1-x} + \sin^{-1} (2\sqrt{x(1-x)})$ હોય,તો $\frac{dy}{dx}$ ની કિંમત શોધો.

$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239}$ ની કિંમત શોધો.

કિંમત શોધો: $\tan ^2(\sec ^{-1} 3) + \operatorname{cosec}^2(\cot ^{-1} 2) + \cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3}) = $ . . . . . . .

જો $\tan ^{-1} a+\tan ^{-1} b+\tan ^{-1} c=\pi$ હોય,તો નીચેનામાંથી કયું સાચું છે?

સાબિત કરો કે $\tan ^{-1} \sqrt{x} = \frac{1}{2} \cos ^{-1} \left( \frac{1-x}{1+x} \right)$,જ્યાં $x \in [0, 1]$.

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