Question

tan theta =1/2 then evaluate cos theta/sintheta +sintheta/1+ cos thet​

Answer 1

Answer:

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Answer 2

Given: $\tan \theta=\frac{1}{2}$

we have to find: $\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{1+\cos \theta} =?$

From given value we get:

opposite=$1$

Adjacent=$2$

Hypotonuse=$\sqrt{1^{2}+2^{2} } =\sqrt{1+4} =\sqrt{5}$

Therefore:

$\cos \theta=\frac{2}{\sqrt{5} }$

$\sin \theta=\frac{1}{\sqrt{5} }$

therefore:

$\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{1+\cos \theta}$

$=\frac{\frac{2}{\sqrt{5} } }{\frac{1}{\sqrt{5} } } +\frac{\frac{1}{\sqrt{5} } }{1+\frac{2}{\sqrt{5} } }$

$=\frac{\frac{2}{\sqrt{5} } }{\frac{1}{\sqrt{5} } } +\frac{\frac{1}{\sqrt{5} } }{\frac{\sqrt{5}+2 }{\sqrt{5} } } }$

$=\frac{2}{\sqrt{5} } \times \frac{\sqrt{5} }{1} +\frac{1}{\sqrt{5} } \times\frac{\sqrt{5} }{\sqrt{5} +2}$

$=2+\frac{1}{\sqrt{5} +2}$

rationalizing:

$=2+\frac{1}{\sqrt{5} +2}\times\frac{ {\sqrt{5} -2}}{ {\sqrt{5} -2}}$

$=2+\frac{\sqrt{5} -2}{\sqrt{5} ^{2}-2^{2} }$

$=2+\frac{\sqrt{5} -2}{5-4}$

$=2+\sqrt{5} -2$

$=\sqrt{5} .$

Therefore:

$\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{1+\cos \theta} =\sqrt{5}.$