Question

given that sin theata = a/b' then tan theata is ​

Answer 1

$\large\bf\underline\red{Answer \: :-}$

Given :

$\sf{ \sin \theta = \frac{a}{b} }$

To find :

tan θ

Now,

We know that

tan θ = sin θ / cos θ

Now we find cos θ

$\sf\longrightarrow{ \cos \theta = \sqrt{1 - \sin {}^{2} \theta } } \\ \\ \sf\longrightarrow{ \cos \theta = \sqrt{1 - \frac{a {}^{2} }{b {}^{2} } } } \: \: \: \\ \\ \sf\longrightarrow{ \cos \theta = \sqrt{ \frac{b {}^{2} - a {}^{2} }{b {}^{2} } } } \: \\ \\ \sf\longrightarrow{ \cos \theta = \sqrt{ \frac{b {}^{2} - a {}^{2} }{b} } }$

Now,

$\sf\bigstar{ \: \tan \theta = \frac{ \sin \theta }{ \cos \theta } }$

Finding tan theta

$\sf\longrightarrow{ \tan \theta = \frac{a/b}{ \sqrt{ \frac{b {}^{2} - a {}^{2} }{b} } } } \: \: \: \: \\ \\ \\ \sf\longrightarrow \pink{ \tan \theta = \frac{a}{ \sqrt{b {}^{2} - a {}^{2} } } }$

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Trigonometry table :-

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