(cos0°+sin45°+sin30°)(sin90°-cos45°+cos60°). Plz give the answer.....plz

We know that, cos0°=1 =sin90°, sin45°=(1/√2)= cos45° & sin30°=(1/2)=cos60°, putting these values in the given expression, we get:-
{tex}\left( {\cos 0^\circ +\sin 45^\circ +\sin 30^\circ }\right)\left( {\sin 90^\circ -\cos 45^\circ +\cos 60^\circ } \right){/tex}
{tex} = \left( {1 + \frac{1}{{\sqrt 2 }} + \frac{1}{2}} \right)\left( {1 - \frac{1}{{\sqrt 2 }} + \frac{1}{2}} \right){/tex}
{tex} = \left( {\frac{{2\sqrt 2 + 2 + \sqrt 2 }}{{2\sqrt 2 }}} \right)\left( {\frac{{2\sqrt 2 - 2 + \sqrt 2 }}{{2\sqrt 2 }}} \right){/tex}
{tex} = \left( {\frac{{3\sqrt 2 + 2}}{{2\sqrt 2 }}} \right)\left( {\frac{{3\sqrt 2 - 2}}{{2\sqrt 2 }}} \right){/tex}
{tex} = \frac{{{{\left( {3\sqrt 2 } \right)}^2} - {{\left( 2 \right)}^2}}}{8}{/tex} [Identity (a + b)(a - b) = a² - b²]
{tex} = \frac{{18 - 4}}{8}{/tex}
{tex} = \frac{{14}}{8} = \frac{7}{4}{/tex}