જો બે ખૂણાઓ alpha, beta એવા હોય કે 0

Question

(D) આપેલ છે,$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ અને $\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}}=\frac{1}{7}$.$1+\cos 2 \alpha = 2 \cos^2 \al

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) આપેલ છે,$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ અને $\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}}=\frac{1}{7}$.$1+\cos 2 \alpha = 2 \cos^2 \alpha$ હોવાથી,$\sqrt{2} \cos \alpha = \frac{3}{\sqrt{5}} \Rightarrow \cos \alpha = \frac{3}{\sqrt{10}}$.તેથી $\cos 2 \alpha = 2 \cos^2 \alpha - 1 = 2(\frac{9}{10}) - 1 = \frac{8}{10} = \frac{4}{5}$.$\sin^2 2 \alpha = 1 - \cos^2 2 \alpha = 1 - \frac{16}{25} = \frac{9}{25}$ હોવાથી,$\sin 2 \alpha = \frac{3}{5}$.આમ,$	an 2 \alpha = \frac{\sin 2 \alpha}{\cos 2 \alpha} = \frac{3/5}{4/5} = \frac{3}{4}$.$\beta$ માટે,$\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}} = 	an \beta = \frac{1}{7}$.હવે,$	an(2 \alpha + \beta) = \frac{	an 2 \alpha + 	an \beta}{1 - 	an 2 \alpha 	an \beta} = \frac{3/4 + 1/7}{1 - (3/4)(1/7)} = \frac{25/28}{25/28} = 1$.તેથી,$2 \alpha + \beta = 	an^{-1}(1) = \frac{\pi}{4}$.

જો બે ખૂણાઓ $\alpha, \beta$ એવા હોય કે $0 < \alpha, \beta < \frac{\pi}{4}$,$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ અને $\frac{\sqrt{1-\cos 2 \beta}}{\sqrt{1+\cos 2 \beta}}=\frac{1}{7}$,તો $(2 \alpha+\beta)=$

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