Question 706383
Use the identity {{{ sinA=2sinAcosA }}}


Then:
{{{ sin2x/sinx = 2sinxcosx/sinx }}}
{{{ sin2x/sinx = 2cosx }}}
{{{ sin2x/sinx = 2/(1/cosx) }}}
{{{ sin2x/sinx = 2/(secx) }}} QED