यदि theta न्यून कोण है तथा sin frac{theta }{ | vedclass

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Question

(b) $	an 	heta = \frac{{\sin 	heta }}{{\cos 	heta }}$

$	an 	heta = \frac{{2\sin \frac{	heta }{2}\cos \frac{	heta }{2}}}{{1 - 2{{\sin }^2}\f

Vedclass · Accepted Answer

$\sqrt {{x^2} - 1} $

Vedclass · Answer

(b) $	an 	heta = \frac{{\sin 	heta }}{{\cos 	heta }}$

$	an 	heta = \frac{{2\sin \frac{	heta }{2}\cos \frac{	heta }{2}}}{{1 - 2{{\sin }^2}\frac{	heta }{2}}} = \frac{{2	an \frac{	heta }{2}}}{{1 - {{	an }^2}\frac{	heta }{2}}}$

$\left[ \begin{array}{l}{m{Using\,}}\,\,{m{sin}}\frac{	heta }{{m{2}}} = \sqrt {\frac{{x - 1}}{{2x}}} \	herefore \,\,\cos \frac{	heta }{2} = \sqrt {1 - {{\sin }^2}\frac{	heta }{2}} = \sqrt {\frac{{x + 1}}{{2x}}} {m{\, \,and\,\,}}\,{m{tan}}\frac{	heta }{{m{2}}} = \frac{{\sqrt {x - 1} }}{{\sqrt {x + 1} }}\end{array} ight]$

$	herefore \,\,\,	an 	heta = \sqrt {{x^2} - 1} $.

यदि $\theta $ न्यून कोण है तथा $\sin \frac{\theta }{2} = \sqrt {\frac{{x - 1}}{{2x}}} $, तो $\tan \theta $ का मान है

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