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

$\tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) = $

$\tan \frac{1}{2} \left[ \sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2} \right]$ ની કિંમત શોધો,જ્યાં $|x | < 1, y>0$ અને $xy < 1$ છે.

$2 \pi - \left(\sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}\right)$ ની કિંમત શોધો.

ધારો કે $S = \{x \in R : 0 < x < 1 \text{ અને } 2 \tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\}$. જો $n(S)$ એ $S$ માં રહેલા ઘટકોની સંખ્યા દર્શાવતું હોય,તો:

સાબિત કરો કે $2 \sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{24}{7}$.

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