operatorname{cosec}left[2 cot ^{-1}(5)+cos ^{ | vedclass

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Question

(B) ધારો કે $x = 2 \cot^{-1}(5) + \cos^{-1}(\frac{4}{5})$.આપણે જાણીએ છીએ કે $\cot^{-1}(5) = 	an^{-1}(\frac{1}{5})$.તેથી,$2 \cot^{-1}(5) = 2 	an^{-1}

Vedclass · Accepted Answer

$\frac{65}{56}$

Vedclass · Answer

(B) ધારો કે $x = 2 \cot^{-1}(5) + \cos^{-1}(\frac{4}{5})$.આપણે જાણીએ છીએ કે $\cot^{-1}(5) = 	an^{-1}(\frac{1}{5})$.તેથી,$2 \cot^{-1}(5) = 2 	an^{-1}(\frac{1}{5}) = 	an^{-1}(\frac{2(1/5)}{1-(1/5)^2}) = 	an^{-1}(\frac{2/5}{24/25}) = 	an^{-1}(\frac{5}{12})$.વળી,$\cos^{-1}(\frac{4}{5}) = 	an^{-1}(\frac{3}{4})$ કારણ કે જો $\cos 	heta = \frac{4}{5}$,તો $	an 	heta = \frac{3}{4}$.આમ,પદાવલિ $\operatorname{cosec}[	an^{-1}(\frac{5}{12}) + 	an^{-1}(\frac{3}{4})]$ બને છે.સૂત્ર $	an^{-1} A + 	an^{-1} B = 	an^{-1}(\frac{A+B}{1-AB})$ નો ઉપયોગ કરતા:$	an^{-1}(\frac{5/12 + 3/4}{1 - (5/12)(3/4)}) = 	an^{-1}(\frac{14/12}{33/48}) = 	an^{-1}(\frac{56}{33})$.અંતે,$\operatorname{cosec}(	an^{-1}(\frac{56}{33}))$. જો $	heta = 	an^{-1}(\frac{56}{33})$,તો $	an 	heta = \frac{56}{33}$.$\operatorname{cosec} 	heta = \sqrt{1 + \cot^2 	heta} = \sqrt{1 + (33/56)^2} = \sqrt{\frac{3136 + 1089}{3136}} = \sqrt{\frac{4225}{3136}} = \frac{65}{56}$.

$\operatorname{cosec}\left[2 \cot ^{-1}(5)+\cos ^{-1}\left(\frac{4}{5}\right)\right]$ ની કિંમત ..... છે.

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