Question

Please solve the question..

Answer 1

GIVEN :–

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• $\bf \tan(A) = \sqrt{2} - 1$

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TO FIND :–

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• $\bf \dfrac{\tan(A)}{1+\tan^2(A)} = ?$

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SOLUTION :–

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$\bf \: \: =\: \: \dfrac{\tan(A)}{1+\tan^2(A)}$

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• Put the values –

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$\bf \: \: =\: \: \dfrac{( \sqrt{2} - 1)}{1+( \sqrt{2} - 1)^2}$

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• Using identity –

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$\bf \: \: \to \: \: {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

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• So that –

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$\bf \: \: =\:\: \dfrac{( \sqrt{2} - 1)}{1+( \sqrt{2}) ^{2} + (1)^2 - 2 \sqrt{2} }$

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$\bf \: \: =\:\: \dfrac{( \sqrt{2} - 1)}{1+2+1- 2 \sqrt{2} }$

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$\bf \: \: =\:\: \dfrac{( \sqrt{2} - 1)}{2+2- 2 \sqrt{2} }$

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$\bf \: \: =\:\: \dfrac{( \sqrt{2} - 1)}{4- 2 \sqrt{2} }$

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$\bf \: \: =\:\: \dfrac{ \cancel{( \sqrt{2} - 1)}}{2 \sqrt{2} \cancel{( \sqrt{2} - 1)}}$

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$\bf \: \: =\:\: \dfrac{1}{2 \sqrt{2}}$

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▪︎ Hence –

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$\bf \implies \large{ \boxed{ \bf \dfrac{\tan(A)}{1+\tan^2(A)}=\:\: \dfrac{1}{2 \sqrt{2}}}}$