sin^{-1} left( frac{12}{13} right) - sin^{-1} | vedclass

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

Question

(D) ધારો કે $\alpha = \sin^{-1} \left( \frac{12}{13} \right)$ અને $\beta = \sin^{-1} \left( \frac{3}{5} \right)$.તેથી $\sin \alpha = \frac{12}{13} \im

Vedclass · Accepted Answer

$\frac{\pi}{2} - \sin^{-1} \left( \frac{56}{65} \right)$

Vedclass · Answer

(D) ધારો કે $\alpha = \sin^{-1} \left( \frac{12}{13} ight)$ અને $\beta = \sin^{-1} \left( \frac{3}{5} ight)$.તેથી $\sin \alpha = \frac{12}{13} \implies \cos \alpha = \sqrt{1 - \left( \frac{12}{13} ight)^2} = \frac{5}{13}$.અને $\sin \beta = \frac{3}{5} \implies \cos \beta = \sqrt{1 - \left( \frac{3}{5} ight)^2} = \frac{4}{5}$.સૂત્ર $\sin^{-1} x - \sin^{-1} y = \sin^{-1} (x \sqrt{1 - y^2} - y \sqrt{1 - x^2})$ નો ઉપયોગ કરતા:$\sin^{-1} \left( \frac{12}{13} ight) - \sin^{-1} \left( \frac{3}{5} ight) = \sin^{-1} \left( \frac{12}{13} 	imes \frac{4}{5} - \frac{3}{5} 	imes \frac{5}{13} ight)$$= \sin^{-1} \left( \frac{48}{65} - \frac{15}{65} ight) = \sin^{-1} \left( \frac{33}{65} ight)$.આપણે જાણીએ છીએ કે $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,તેથી $\sin^{-1} \left( \frac{33}{65} ight) = \frac{\pi}{2} - \cos^{-1} \left( \frac{33}{65} ight)$.વૈકલ્પિક રીતે,$\cos^{-1} x = \sin^{-1} \sqrt{1 - x^2}$ નો ઉપયોગ કરતા,$\sin^{-1} \left( \frac{33}{65} ight) = \cos^{-1} \sqrt{1 - \left( \frac{33}{65} ight)^2} = \cos^{-1} \left( \frac{56}{65} ight) = \frac{\pi}{2} - \sin^{-1} \left( \frac{56}{65} ight)$.આમ,સાચો વિકલ્પ $D$ છે.

$\sin^{-1} \left( \frac{12}{13} \right) - \sin^{-1} \left( \frac{3}{5} \right)$ નું મૂલ્ય કેટલું થાય?

Similar Questions

$\tan \left[ 2\tan^{-1}\left( \frac{1}{5} \right) - \frac{\pi}{4} \right] = $

વિધેયને તેના સરળ સ્વરૂપમાં લખો: $\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right), a>0 ; \frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}$

કિંમત શોધો: $\tan^{-1} \left( \frac{a - b}{1 + ab} \right) + \tan^{-1} \left( \frac{b - c}{1 + bc} \right)$

સાબિત કરો કે $\sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{84}{85}$

$\tan \left[ \cos^{-1} \frac{4}{5} + \tan^{-1} \frac{2}{3} \right] =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module