Question

Evaluate the following :-
1).
$\sin( {60}^{0 } ) \cos( {30}^{0} ) + \sin( {30}^{0} ) \cos( {60}^{0} )$
2).
$2 \: ta{n}^{2} {45}^{0} + co{s}^{2} {30}^{0} - si{n}^{2} {60}^{0}$
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Answer 1

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\star\: {\bf{ sin60 \degree \: cos 30 \degree + sin30 \degree \: cos60\degree}}$

$\star\: {\bf{ \: 2 {tan}^{2} 45 \degree + {cos}^{2} 30 \degree - sin^{2} 60 \degree }}$

${\bf{\blue{\underline{Now,}}}}$

$(1).\star\: {\bf{ sin60 \degree \: cos 30 \degree + sin30 \degree \: cos60\degree}}$

We know that

${ \boxed {\bf{ \odot \: sin \: 60 \degree = \frac{ \sqrt{3} }{2} }}}$

${ \boxed {\bf{ \odot \: cos \: 60 \degree = \frac{ \sqrt{3} }{2} }}}$

${ \boxed {\bf{ \odot \: sin \: 30 \degree = \frac{ 1}{2} }}}$

${ \boxed {\bf{ \odot \: cos \: 60 \degree = \frac{ 1}{2} }}}$

Now put all values in given equation,

$\implies{\bf{ \big( \frac{ \sqrt{3} }{2} \big) \times \big( \frac{ \sqrt{3} }{2} \big) + \frac{1}{2} \times \frac{1}{2} }}$

$\implies{\bf{ \frac{3}{4} + \frac{1}{4} }}$

$\implies{\bf{ \frac{3+1}{4} }}$

$\implies{\bf{ \frac{4}{4} }}$

$\huge { \boxed{\bf{ { \purple {\: 1 \: Ans }}}}}$

Now,

$(2.)\star\: {\bf{ \: 2 {tan}^{2} 45 \degree + {cos}^{2} 30 \degree - sin^{2} 60 \degree }}$

${ \boxed {\bf{ \odot \: tan \: 45 \degree = 1 }}}$

${ \boxed {\bf{ \odot \: cos \: 30\degree = \frac{ \sqrt{3} }{2} }}}$

${ \boxed {\bf{ \odot \: sin \: 60 \degree = \frac{ \sqrt{3} }{2} }}}$

Now put all values in given equation,

$\implies{\bf{ 2 {(1)}^{2} + \big( \frac{ \sqrt{3} }{2} \big) ^{2} - \big( \frac{ \sqrt{3} }{2} \big) ^{2} }}$

$\implies{\bf{ 2(1) + \frac{3}{4} - \frac{3}{4} }}$

$\implies{\bf{ 2(1) + \frac{3 - 3}{4} }}$

$\implies{\bf{ 2(1) + \frac{0}{4} }}$

$\implies{\bf{ 2(1) + 0 }}$

$\huge { \boxed{\bf{ { \purple {\: 2 \: Ans }}}}}$

Answer 2

Given :-

$\bullet\sf sin60^{\circ} . cos30^{\circ} + sin30^{\circ}. cos60^{\circ}\\ \\ \bullet\sf 2 tan^{2} 45^{\circ} + cos^2 30^{\circ} -sin^2 60^{\circ}$

We know that :-

$\sf \star\ \ sin60^{\circ}= \dfrac{\sqrt{3}}{2}\ ; \ \sf \star\ cos 30^{\circ}= \dfrac{\sqrt{3}}{2}\\ \\ \sf \star \ sin30^{\circ}= \dfrac{1}{2}\ ; \ \sf \star \ cos\ 60^{\circ}= \dfrac{1}{2}$

Now put the values in Q1)

$:\implies\sf sin60^{\circ} . cos30^{\circ} + sin30^{\circ}. cos60^{\circ}\\ \\ :\implies\sf \bigg[ \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2} \bigg] + \bigg[\dfrac{1}{2} \times \dfrac{1}{2}\bigg]\\ \\ :\implies\sf \bigg[ \dfrac{\sqrt{3}\times \sqrt{3}}{2\times 2} \bigg] +\bigg[ \dfrac{1}{2\times 2}\bigg]\\ \\ :\implies\sf \bigg[ \dfrac{3}{4}+\dfrac{1}{4}\bigg] \\ \\ :\implies\sf \dfrac{3+1}{4} = \cancel{\dfrac{4}{4}}\\ \\ :\implies\sf 1$

$\underline{\boxed{\sf\ Answer :\ 1}}$

Now Q-2

$:\implies\sf 2 tan^{2} 45^{\circ} + cos^2 30^{\circ} -sin^2 60^{\circ}\\ \\ \sf \star \ tan45^{\circ}= 1\ ;\ \star\ cos 30^{\circ}= \dfrac{\sqrt{3}}{2}\ \ ;\ \star\ sin60^{\circ}= \dfrac{\sqrt{3}}{2}$

$:\implies\sf 2 tan^{2} 45^{\circ} + cos^2 30^{\circ} -sin^2 60^{\circ}\\ \\ \sf \ put \ the \ values \\ \\ :\implies\sf ( 2\times 1) +\bigg[\dfrac{\sqrt{3}}{2}\bigg]^2 - \bigg[\dfrac{\sqrt{3}}{2}\bigg]^2\\ \\ :\implies\sf 2+ \dfrac{3}{4}-\dfrac{3}{4}\\ \\ :\implies\sf \bigg[\dfrac{8+\cancel{3}-\cancel{3}}{4}\bigg]\\ \\ :\implies\sf \cancel{\dfrac{8}{4}}\\ \\ :\implies\sf 2$

$\underline{\boxed{\sf\ Answer:\ 2}}$