Question

13) Prove that, 1−tan 45 / 1+tan 45

= 0.​

Answer 1

We have to prove $\sf \dfrac{1 - tan\;45^\circ}{1 + tan\;45^\circ} = 0$.

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We know that,

$\sf tan\;\theta = \dfrac{sin\;\theta}{cos\;\theta}$

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$\therefore\;\sf tan\;45^\circ = \dfrac{sin\;45^\circ}{cos\;45^\circ}$

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$\dashrightarrow\sf tan\;45^\circ = \dfrac{ \frac{1}{ \sqrt{2}}}{ \frac{1}{ \sqrt{2}}}\;\;\;\;\;\;\bigg\lgroup\bf sin\;45^\circ\;and\;cos\;45^\circ = \dfrac{1}{\sqrt{2}} \bigg\rgroup$

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$\dashrightarrow\sf tan\;45^\circ = \dfrac{ \cancel{ \frac{1}{ \sqrt{2}}}}{ \cancel{ \frac{1}{ \sqrt{2}}}}$

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$\dashrightarrow\sf tan\;45^\circ = \bf{1}$

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Therefore,

$:\implies\sf \dfrac{1 - tan\;45^\circ}{1 + tan\;45^\circ}$

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$\dag\;{\underline{\frak{Putting\;value\;of\;tan\;45^\circ}}}$

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$:\implies\sf \dfrac{1 - 1}{1 + 1}$

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$:\implies\sf \dfrac{0}{2}$

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$:\implies{\underline{\boxed{\sf{\purple{\;0\;}}}}}\;\;\;\;\;\;\bigg\lgroup\bf \dfrac{0}{n} = 0\bigg\rgroup$

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Hence ProvEd!

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$\therefore\;\sf \dfrac{1 - tan\;45^\circ}{1 + tan\;45^\circ} = 0$

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★ Trigonometric table:

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$