Question

pls solve this
PT Is prove that
ST is say that​

Answer 1

GIVEN :–

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• $\bf \tan(A) = \dfrac{1}{ \sqrt{3} }$

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TO PROVE :–

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$\bf \to \: 7\sin^{2} (A) + 3 \cos ^{2} (A) = 4$

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SOLUTION :–

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• We know that if –

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$\: \: \: {\huge{.}} \: \: \: \bf \tan(A) = \dfrac{a}{b}$

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• Then –

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$\: \: \: {\huge{.}} \: \: \: \bf \sin(A) = \dfrac{a}{ \sqrt{ {a}^{2} + {b}^{2} } }$

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$\: \: \: {\huge{.}} \: \: \: \bf \cos(A) = \dfrac{b}{ \sqrt{ {a}^{2} + {b}^{2} } }$

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• Let's find sin(A) –

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$\bf \implies \sin(A) = \dfrac{1}{ \sqrt{ {(1)}^{2} + {( \sqrt{3})}^{2} } }$

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$\bf \implies \sin(A) = \dfrac{1}{ \sqrt{1 + 3} }$

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$\bf \implies \sin(A) = \dfrac{1}{ \sqrt{4} }$

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$\bf \implies \large{ \boxed{ \bf \sin(A) = \dfrac{1}{2}}}$

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• Now cos(A) –

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$\bf \implies \sin(A) = \dfrac{ \sqrt{3} }{ \sqrt{ {(1)}^{2} + {( \sqrt{3})}^{2} } }$

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$\bf \implies \sin(A) = \dfrac{ \sqrt{3} }{ \sqrt{1 + 3} }$

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$\bf \implies \sin(A) = \dfrac{ \sqrt{3} }{ \sqrt{4} }$

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$\bf \implies \large{ \boxed{ \bf \sin(A) = \dfrac{ \sqrt{3}} {2}}}$

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• Take L.H.S. –

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$\bf \: \: = \: \: 7\sin^{2} (A) + 3 \cos ^{2} (A)$

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• Put the values –

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$\bf \: \: = \: \: 7 \left( \dfrac{1}{2} \right)^{2} + 3 \left( \dfrac{ \sqrt{3} }{2} \right)^{2}$

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$\bf \: \: = \: \: 7 \left( \dfrac{1}{4} \right) + 3 \left( \dfrac{3}{4} \right)$

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$\bf \: \: = \: \: \dfrac{7}{4} + \dfrac{9}{4}$

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$\bf \: \: = \: \: \dfrac{7 + 9}{4}$

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$\bf \: \: = \: \: \cancel \dfrac{16}{4}$

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$\bf \: \: = \: \: 4$

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$\bf \: \: = \: \: R.H.S.$

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$\bf \: \: \: \: { \underbrace{ \bf Hence \: \: proved}}$