Math, asked by Elsa373, 1 year ago

cos^2 20°+cos^2 70°÷sin^2 59°+sin^2 31°


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Answers

Answered by MaheswariS
22

\textbf{Given:}

\dfrac{cos^220^{\circ}+cos^270^{\circ}}{sin^259^{\circ}+sin^231^{\circ}}

\textbf{To find:}

\text{The value of $\dfrac{cos^220^{\circ}+cos^270^{\circ}}{sin^259^{\circ}+sin^231^{\circ}}$}

\textbf{Solution:}

\text{Using the following results,}

\boxed{\bf\,cos\theta=sin(90^{\circ}-\theta)}

\boxed{\bf\,sin\theta=cos(90^{\circ}-\theta)}

\implies\,cos\,70^{\circ}=sin\,20^{\circ}

sin\,31^{\circ}=cos\,59^{\circ}

\text{Consider,}

\dfrac{cos^220^{\circ}+cos^270^{\circ}}{sin^259^{\circ}+sin^231^{\circ}}

=\dfrac{cos^220^{\circ}+sin^220^{\circ}}{sin^259^{\circ}+cos^259^{\circ}}

\text{Using,}\;\boxed{\bf\,cos^2A+sin^2A=1}

=\dfrac{1}{1}

=1

\therefore\bf\,\dfrac{cos^220^{\circ}+cos^270^{\circ}}{sin^259^{\circ}+sin^231^{\circ}}=1

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Answered by umitbarman1111
15

Answer:

we know that Sin^2 A + Cos^2 A = 1

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