Math, asked by singharshnoor28, 6 hours ago

sin²30-cos²30 +tan²45​

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Answered by navnitkrishnabc663
1

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Answered by mathdude500
6

\large\underline{\sf{Solution-}}

Given Trigonometric expression is

\rm :\longmapsto\: {sin}^{2}30\degree  -  {cos}^{2}30\degree  +  {tan}^{2}45\degree

\rm \:  =  \:  {(sin30\degree )}^{2}  -  {(cos30\degree )}^{2} +  {(tan45\degree )}^{2}

We know, From Trigonometric table of some standard angles,

 \purple{\rm :\longmapsto\:sin30\degree  =  \dfrac{1}{2}}

 \purple{\rm :\longmapsto\:cos30\degree  =  \dfrac{ \sqrt{3} }{2}}

 \purple{\rm :\longmapsto\:tan45\degree  = 1}

So, on substituting these values, we get

\rm \:  =  \:  {\bigg[\dfrac{1}{2} \bigg]}^{2}  -  {\bigg[\dfrac{ \sqrt{3} }{2} \bigg]}^{2} +  {(1)}^{2}

\rm \:  =  \: \dfrac{1}{4}  - \dfrac{3}{4}  + 1

\rm \:  =  \: \dfrac{1 - 3 + 4}{4}

\rm \:  =  \: \dfrac{5 - 3}{4}

\rm \:  =  \: \dfrac{2}{4}

\rm \:  =  \: \dfrac{1}{2}

Hence,

 \purple{\rm\implies \:\boxed{\tt{ \: {sin}^{2}30\degree  -  {cos}^{2}30\degree  +  {tan}^{2}45\degree  =  \frac{1}{2} \: }}}

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MORE TO KNOW

\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

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