Question

evalute
$\frac{2 cos²60⁰ + 3 sec²30⁰ - tan²45⁰}{sin²45⁰ + cos²45⁰}$
​

Answer 1

TO FIND :–

• The value of :–

$\\\implies \bf\dfrac{2 cos^{2}60^{\circ} + 3 sec^{2}30^{\circ} - tan^{2}45^{\circ}}{sin^{2}45^{\circ}+ cos^{2}45^{\circ}} = ?$

SOLUTION :–

• Let –

$\\ \bf \implies P = \dfrac{2 cos^{2}60^{\circ}+ 3 sec^{2}30^{\circ} - tan^{2}45^{\circ}}{sin^{2}45^{\circ}+ cos^{2}45^{\circ}}$

• We know that –

$\\ \bf { \huge{.}} \: \cos( {60}^{ \circ} ) = \dfrac{1}{2}$

$\\ \bf { \huge{.}} \: \sec( {30}^{ \circ} ) = \dfrac{2}{ \sqrt{3} }$

$\\ \bf { \huge{.}} \: \tan( {45}^{ \circ} ) =1$

$\\ \bf { \huge{.}} \: \sin( {45}^{ \circ} ) = \dfrac{1}{ \sqrt{2} }$

$\\ \bf { \huge{.}} \: \cos( {45}^{ \circ} ) = \dfrac{1}{ \sqrt{2} }$

• Now put the values –

$\\ \bf \implies P = \dfrac{2 { \bigg( \dfrac{1}{2} \bigg)}^{2} + 3 { \bigg( \dfrac{2}{ \sqrt{3} } \bigg)}^{2} - 1}{{ \bigg( \dfrac{1}{ \sqrt{2} } \bigg)}^{2}+ { \bigg( \dfrac{1}{ \sqrt{2} } \bigg)}^{2}}$

$\\ \bf \implies P = \dfrac{2 \times {\dfrac{1}{4}}+ 3 \times \dfrac{4}{3} - 1}{ \dfrac{1}{2} +\dfrac{1}{2}}$

$\\ \bf \implies P = \dfrac{{\dfrac{1}{2}}+ 4- 1}{1}$

$\\ \bf \implies P = {\dfrac{1}{2}}+3$

$\\ \bf \implies P = {\dfrac{1 + 6}{2}}$

$\\ \large \implies{ \boxed{ \bf P = {\dfrac{7}{2}}}}$

Answer 2

${\pink{To \: Find:-}}$

• The value of :–

$\begin{gathered}\\\implies \bf\dfrac{2 cos^{2}60^{\circ} + 3 sec^{2}30^{\circ} - tan^{2}45^{\circ}}{sin^{2}45^{\circ}+ cos^{2}45^{\circ}} = ?\\\end{gathered}$

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

• Let:-

$\begin{gathered} \\ \bf \implies P = \dfrac{2 cos^{2}60^{\circ}+ 3 sec^{2}30^{\circ} - tan^{2}45^{\circ}}{sin^{2}45^{\circ}+ cos^{2}45^{\circ}} \\ \end{gathered}$

• We know that:-

$\begin{gathered} \\ \bf { \huge{.}} \: \cos( {60}^{ \circ} ) = \dfrac{1}{2} \\ \end{gathered}$

$\begin{gathered} \\ \bf { \huge{.}} \: \sec( {30}^{ \circ} ) = \dfrac{2}{ \sqrt{3} } \\ \end{gathered}$

$\begin{gathered} \\ \bf { \huge{.}} \: \tan( {45}^{ \circ} ) =1\\ \end{gathered}$

$\begin{gathered} \\ \bf { \huge{.}} \: \sin( {45}^{ \circ} ) = \dfrac{1}{ \sqrt{2} } \\ \end{gathered}$

$\begin{gathered} \\ \bf { \huge{.}} \: \cos( {45}^{ \circ} ) = \dfrac{1}{ \sqrt{2} } \\ \end{gathered}$

• Now put the values:-

$\begin{gathered} \\ \bf \implies P = \dfrac{2 { \bigg( \dfrac{1}{2} \bigg)}^{2} + 3 { \bigg( \dfrac{2}{ \sqrt{3} } \bigg)}^{2} - 1}{{ \bigg( \dfrac{1}{ \sqrt{2} } \bigg)}^{2}+ { \bigg( \dfrac{1}{ \sqrt{2} } \bigg)}^{2}} \\ \end{gathered}$

$\begin{gathered} \\ \bf \implies P = \dfrac{2 \times {\dfrac{1}{4}}+ 3 \times \dfrac{4}{3} - 1}{ \dfrac{1}{2} +\dfrac{1}{2}} \\ \end{gathered}$

$\begin{gathered} \\ \bf \implies P = \dfrac{{\dfrac{1}{2}}+ 4- 1}{1} \\ \end{gathered}$

$\begin{gathered} \\ \bf \implies P = {\dfrac{1}{2}}+3\\ \end{gathered}$

$\begin{gathered} \\ \bf \implies P = {\dfrac{1 + 6}{2}}\\ \end{gathered}$

$\begin{gathered} \\ \large \implies{ \boxed{ \bf P = {\dfrac{7}{2}}}}\\ \end{gathered}$

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