જો tan alpha = frac{1}{7} અને tan beta = frac | vedclass

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(B) આપેલ છે $	an \alpha = \frac{1}{7}$.સૂત્ર $\cos 2\alpha = \frac{1 - 	an^2 \alpha}{1 + 	an^2 \alpha}$ નો ઉપયોગ કરતા:$\cos 2\alpha = \frac{1 - (1/

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$\sin 4\beta$

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(B) આપેલ છે $	an \alpha = \frac{1}{7}$.સૂત્ર $\cos 2\alpha = \frac{1 - 	an^2 \alpha}{1 + 	an^2 \alpha}$ નો ઉપયોગ કરતા:$\cos 2\alpha = \frac{1 - (1/7)^2}{1 + (1/7)^2} = \frac{1 - 1/49}{1 + 1/49} = \frac{48/49}{50/49} = \frac{48}{50} = \frac{24}{25}$.હવે,આપેલ છે $	an \beta = \frac{1}{3}$.સૂત્ર $\sin 2\beta = \frac{2 	an \beta}{1 + 	an^2 \beta}$ નો ઉપયોગ કરતા:$\sin 2\beta = \frac{2(1/3)}{1 + (1/3)^2} = \frac{2/3}{1 + 1/9} = \frac{2/3}{10/9} = \frac{2}{3} 	imes \frac{9}{10} = \frac{3}{5}$.$\sin 2\beta = \frac{3}{5}$ હોવાથી,$\cos 2\beta = \sqrt{1 - (3/5)^2} = \frac{4}{5}$.હવે,$\sin 4\beta = 2 \sin 2\beta \cos 2\beta = 2 	imes \frac{3}{5} 	imes \frac{4}{5} = \frac{24}{25}$.આમ,$\cos 2\alpha = \sin 4\beta$.

જો $\tan \alpha = \frac{1}{7}$ અને $\tan \beta = \frac{1}{3}$ હોય,તો $\cos 2\alpha = $

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