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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Trignometry

Find the value of $\tan 22\large\frac{1}{2}^{\large\circ}$

$(a)\;\sqrt 2\qquad(b)\;\sqrt 2+2\qquad(c)\;2\sqrt 2\qquad(d)\;\sqrt 2-1$

1 Answer

We know that
$\tan\theta=\sqrt{\large\frac{1-\cos 2\theta}{1+\cos 2\theta}}$
$\tan 22\large\frac{1}{2}^{\large\circ}=\sqrt{\large\frac{1-\cos 45^{\large\circ}}{1+\cos 45^{\large\circ}}}$
$\qquad\qquad=\sqrt{\large\frac{1-\Large\frac{1}{\sqrt 2}}{1+\Large\frac{1}{\sqrt 2}}}$
$\qquad\qquad=\sqrt{\large\frac{\sqrt 2-1}{\sqrt 2+1}}$
$\sqrt{\large\frac{\sqrt 2-1}{\sqrt 2+1}\times \sqrt{\large\frac{\sqrt 2-1}{\sqrt 2-1}}}=\sqrt{\large\frac{(\sqrt 2-1)^2}{(\sqrt 2+1)(\sqrt 2-1)}}$
$\Rightarrow \sqrt{\large\frac{(\sqrt 2-1)^2}{2-1}}$
$\Rightarrow \sqrt{(\sqrt 2-1)^2}$
$\Rightarrow \sqrt 2-1$
Hence (d) is the correct answer.
answered Oct 9, 2013 by sreemathi.v
 

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