Sunday, May 19, 2019

IGNOU MTE01 2019 SOLVED ASSIGNMENT 1(a)(iii)

Put  x = \tan \theta
+
Thus, \theta = {\tan ^{ - 1}}x

y = \sqrt {\frac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}}

y = \sqrt {\sec 2\theta }


\frac{{dy}}{{dx}} = \frac{{\sec 2\theta *\tan 2\theta }}{{\sqrt {\sec 2\theta } }}

\frac{{dy}}{{dx}} = \tan 2\theta \sqrt {\sec 2\theta }

\frac{{dy}}{{dx}} = \frac{{\sin 2\theta }}{{\sqrt {{{\left( {\cos 2\theta } \right)}^3}} }}

Replacing


\sin 2\theta = \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }},and


\cos 2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}

in


\frac{{dy}}{{dx}} = \frac{{\sin 2\theta }}{{\sqrt {{{\left( {\cos 2\theta } \right)}^3}} }}

and solving we get

\frac{{dy}}{{dx}} = \frac{{2\tan \theta *\sqrt {1 + {{\tan }^2}\theta } }}{{1 - {{\tan }^2}\theta \sqrt {1 - {{\tan }^2}\theta } }}

Substituting back in the above equation x = \tan \theta

\frac{{dy}}{{dx}} = \frac{{2x\sqrt {1 + {x^2}} }}{{\sqrt {{{\left( {1 - {x^2}} \right)}^3}} }}

The answer is

\frac{{dy}}{{dx}} = \frac{{2x\sqrt {1 + {x^2}} }}{{\left( {1 - {x^2}} \right)\sqrt {\left( {1 - {x^2}} \right)} }}



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