Question

if tan a + sec a =2/root 3 find the value of sin a ​

Answer 1

Answer:

$\frac{1}{7}$

Step-by-step explanation:

Given,

$\sec( \alpha ) + \tan( \alpha ) = \frac{2}{ \sqrt{3} } ..........(i)$

$= \frac{1}{ \sec( \alpha ) - \tan( \alpha ) } = \frac{2}{ \sqrt{3} }$

$= \frac{ \sec( \alpha ) + \tan( \alpha ) }{ ({ \sec( \alpha )) }^{2} - ( \tan( \alpha ))^{2} } = \frac{2}{ \sqrt{3} }$

$= \frac{ \sec( \alpha ) + \tan( \alpha ) }{( \sec( \alpha ) + \tan( \alpha )) \times ( \sec( \alpha ) - \tan( \alpha )) } = \frac{2}{ \sqrt{3} }$

$= \sec( \alpha ) - \tan( \alpha ) = \frac{ \sqrt{3} }{2} .....(ii)$

Now, adding (i) and (ii), we get

$= sec( \alpha ) = \frac{7}{4 \sqrt{3} }$

$= \cos( \alpha ) = \frac{4 \sqrt{3} }{7}$

so,

$\sin( \alpha ) = \sqrt{1 - ( \cos( \alpha ) )^{2} }$

=

$\frac{1}{7}$