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If\ \sin x +\cos x=\frac{\sqrt{5}}{2} Find the value of \tan(\frac{x}{2}) ?

Option: 1

\frac{2\pm \sqrt{12}}{2(\sqrt{5}+2)}


Option: 2

\frac{4- \sqrt{12}}{2(\sqrt{5}+2)}


Option: 3

\frac{4+ \sqrt{12}}{2(\sqrt{5}+2)}


Option: 4

\frac{4\pm \sqrt{12}}{2(\sqrt{5}+2)}


Answers (1)

best_answer

Trigonometric Ratio of Submultiple of an Angle -

Trigonometric Ratio of Submultiple of an Angle

 

Trigonometric Ratio of θ in terms of θ/2 
\begin{aligned} \sin ( \theta) &=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \\&=\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}} \\\cos ( \theta) &=\cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2} \\ &=1-2 \sin ^{2} \frac{\theta}{2} \\ &=2 \cos ^{2} \frac{\theta}{2}-1\\&=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}} \\ \tan ( \theta) &=\frac{2 \tan \frac{\theta}{2}}{1-\tan ^{2} \frac{\theta}{2}} \end{aligned}

 

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\sin x +\cos x=\frac{\sqrt{5}}{2}\\ \frac{2\tan \frac {x}{2}}{1+ \tan^2 \frac{x}{2}}+\frac{1-\tan^2 \frac {x}{2}}{1+ \tan^2 \frac{x}{2}}=\frac{\sqrt{5}}{2}\\ \frac{2 \tan \frac {x}{2}+1- \tan^2 \frac{x}{2}}{1+\tan^2 \frac {x}{2}}=\frac{\sqrt{5}}{2}\\ (\sqrt{5}+2)\tan^2 \frac {x}{2}-4 \tan \frac{x}{2}+\sqrt{5}-2=0\\ \tan \frac{x}{2}= \frac{4\pm \sqrt{16-4*(\sqrt{5}+2)*(\sqrt{5}-2})}{2(\sqrt{5}+2)}\\= \frac{4\pm \sqrt{12}}{2(\sqrt{5}+2)}\sin x +\cos x=\frac{\sqrt{5}}{2}\\ \frac{2\tan \frac {x}{2}}{1+ \tan^2 \frac{x}{2}}+\frac{1-\tan^2 \frac {x}{2}}{1+ \tan^2 \frac{x}{2}}=\frac{\sqrt{5}}{2}\\ \frac{2 \tan \frac {x}{2}+1- \tan^2 \frac{x}{2}}{1+\tan^2 \frac {x}{2}}=\frac{\sqrt{5}}{2}\\ (\sqrt{5}+2)\tan^2 \frac {x}{2}-4 \tan \frac{x}{2}+\sqrt{5}-2=0\\ \tan \frac{x}{2}= \frac{4\pm \sqrt{16-4*(\sqrt{5}+2)*(\sqrt{5}-2})}{2(\sqrt{5}+2)}\\= \frac{4\pm \sqrt{12}}{2(\sqrt{5}+2)}

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Rishabh

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