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  • If <img alt="\frac{\sqrt{2}\sin \alpha }{\sqrt{1+\cos 2\alpha }}=\frac{1}{7}" src="https://learn.careers360.com/latex-image/?%5Cfrac%7B%5Csqrt%7B2%7D%5Csin%20%5Calpha%20%7D%7B%5Csqrt%7B1&plus;%5Ccos%202%5Calpha%20%7D%7D%3D%5Cfrac%7B1%7D%7B7%7D

If \frac{\sqrt{2}\sin \alpha }{\sqrt{1+\cos 2\alpha }}=\frac{1}{7} and \sqrt{\frac{1-\cos 2\beta }{2}}=\frac{1}{\sqrt{10}}, \alpha ,\beta\; \epsilon \left ( 0,\frac{\pi }{2} \right ), then \tan \left ( \alpha +2\beta \right ) is equal to ________.
Option: 1 0  
Option: 2 1
Option: 3 0.5
Option: 4 2
 

Answers (1)

best_answer

 

 

Trigonometric Identities -

Trigonometric Identities-

These identities are the equations that hold true regardless of the angle being chosen.

 

\\\mathrm{\sin^2\mathit{t}+\cos^2\mathit{t}=1}\\\mathrm{1+\tan^2\mathit{t}=\sec^2\mathit{t}}\\\mathrm1+{\cot^2\mathit{t}=\csc^2\mathit{t}}\\\mathrm{\tan \mathit{t}=\frac{\sin \mathit{t}}{\cos \mathit{t}},\;\;\cot \mathit{t}=\frac{\cos\mathit{t}}{\sin\mathit{t}}}

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Double Angle Formula and Reduction Formula -

Double Angle Formula and Reduction Formula

\begin{aligned} \sin (2 \theta) &=2 \sin \theta \cos \theta \\&=\frac{2\tan\theta}{1+\tan^2\theta} \\\cos (2 \theta) &=\cos ^{2} \theta-\sin ^{2} \theta \\ &=1-2 \sin ^{2} \theta \\ &=2 \cos ^{2} \theta-1\\&=\frac{1-\tan^2\theta}{1+\tan^2\theta} \\ \tan (2 \theta) &=\frac{2 \tan \theta}{1-\tan ^{2} \theta} \end{aligned}

 

\begin{aligned} \tan (\alpha+\beta) &=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ \tan (\theta+\theta) &=\frac{\tan \theta+\tan \theta}{1-\tan \theta \tan \theta} \\ \tan (2 \theta) &=\frac{2 \tan \theta}{1-\tan ^{2} \theta} \end{aligned}

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\\\frac{\sqrt{2} \sin \alpha}{\sqrt{2} \cos \alpha}=\frac{1}{7}\\\tan\alpha=\frac{1}{7}\\\sin\beta=\frac{1}{\sqrt{10}}\\\tan\beta=\frac{1}{\sqrt{3}}\\\tan 2 \beta=\frac{2 \cdot \frac{1}{3}}{1-\frac{1}{9}}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{3}{4}\\\tan (\alpha+2 \beta)=\frac{\tan \alpha+\tan 2 \beta}{1-\tan \alpha \tan 2 \beta}=1

Correct Option (2)

Posted by

vishal kumar

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