Question

if cos theta is equal to root 3 upon 2 find the value of 1 + tan theta upon 1 + cosec theta​

Answer 1

$\cos( \theta) = \sqrt{3} \div 2 = \cos(30) \degree \\ \theta = 30\degree \\ (1 + \tan( \theta)) \div (1 + \csc(\theta) ) = (1 + \tan(30) \degree) + (1 + \csc(30) \degree) \\ = (1 +1 \div \sqrt{3}) \div (1 + 2) \\ = (\sqrt{3} + 1) \div ( \sqrt{3} \times 3) \\ = \frac{ \sqrt{3} + 1 }{3 \sqrt{3} } \\ = \frac{3 + \sqrt{3} }{9} \\ = \frac{4.732}{9} \\ = 0.52578$

Answer 2

Given:-

$\sf \implies cos = \frac{ \sqrt{3} }{2}$

To Find:-

To find the value of

$\frac{1 - \sec( \theta) }{1 + \csc( \theta) }$

Solution:-

$\cos( \theta) = \frac{ \sqrt{3} }{2} \\ \\ \sf \cos( \theta) = \cos(30 \degree) \\ \\ \sf \theta = 30 \degree \\ \\ \sf \frac{1 - \sec( \theta) }{1 + \csc( \theta) } = \frac{1 - \sec(30 \degree) }{1 + \csc(30 \degree) }$

As we know that,

$\sf \cos( \theta) = \frac{1}{ \sec( \theta) } \\ \\ \sf \sec(30 \degree) = \frac{2}{ \sqrt{3} } \\ \\ \sf \csc( \theta) = \frac{1}{ \sin( \theta) } \\ \\ \sf \csc(30 \degree) = 2$

$\sf \implies \frac{1 - \frac{2}{ \sqrt{3} } }{1 + 2} \\ \\ \sf \implies \frac{ \frac{ \sqrt{3} - 2}{ \sqrt{3} } }{3} \\ \\ \sf \implies \frac{ \sqrt{3} - 2 }{3 \sqrt{3} }$

By rationalisation,

$\sf \implies \frac{ \sqrt{3} - 2 }{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ \sf \implies \frac{3 - 2 \sqrt{3} }{9}$