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

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

Posted by Ayu Gupta 5 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 5 years, 7 months ago

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

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