Question

tan A=cot A then find cos A​

Answer 1

Answer:

$\tan(a) = \cot(a) \\ \\ \frac{ \sin(a) }{ \cos(a) } = \frac{ \cos(a) }{ \sin(a) } \\ \\ by \: cross \: multiplication - \\ \\ we \: get - \\ { \sin }^{2}a = { \cos }^{2} a \\ \\ \cos(a) = \sqrt{ \sin {}^{2}a }. \\ \\ \\ hope \: it \: will \: help \: u \: dear...$

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Answer 2

GIVEN:

• Tan A= Cot A

TO FIND:

• Cos A

SOLUTION:

$\large \sf : \implies \: \: Tan \: A= Cot \: A$

$\large : \implies \: \: \sf Tan \: A= \frac{1}{Tan \: A}$

$\large \sf : \implies \: \: {Tan }^{2} A = 1$

Cancel Square both sides

$\sf \large : \implies \: \: Tan \: A =1$

$\sf \large : \implies \: \: Tan \: 45 \degree =1$

$\sf So \: A = 45 \degree$

It means we have to find value of Cos 45° :D

$\large : \implies \: \: \sf Cos \: 45 \degree = \frac{1}{ \sqrt{2} }$

$\sf \large: \implies \: \: Cos\:45°= \frac{1 \times \sqrt{2} }{ \sqrt{2} \times \sqrt{2} }$

$\sf \large: \implies \: \: Cos\:45 = \frac{ \sqrt{2} }{2}$

$\sf \large: \implies \: \: Cos\:45 = \frac{ 1.414 }{2}$

$\sf \large: \implies \: \: Cos\:45 = 0.707$

So final answer is 0.707

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