Question

If cosec theeta is 5/3find value of tan theeta

Answer 1

Given cos theta=5/3
Cos theta =adj/hyp =5/3
By Pythagoreas theorm

AC^2=AB^2+BC^2
3^2=5^2+x^2
x^2=25-9
x=root16
Thus,x=4

Tan theta = opp/adj
=4/5

Answer 2

Given:

• Cosec@ = Hypotenuse/Perpendicular

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To Find→

• Value of Tan@

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Solution:-

$\implies Cosec@ = \dfrac{Hypotenuse}{Perpendicular} = \dfrac{5}{3}$

Now Applying Pythagoras theorem:

$\implies {base}^{2} = {Hypotenuse}^{2} - {Perpendicular}^{2}$

$\implies base = \sqrt{{Hypotenuse}^{2} - {Perpendicular}^{2} }$

$\implies \: base = \sqrt{ {5}^{2} - {3}^{2} }$

$\implies base = \sqrt{16}$

$\implies base = 4$

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Tan@ = Perpendicular/base

$\dfrac{3}{4} \: \: \: \red{answer}$

$\bullet \sf Trigonometric Values \\\\\\ \boxed{\begin{array}{c|c|c|c|c|c} \sf Angle & 0^{\circ} & 30^{\circ} & 45^{\circ} & 60^{\circ} & 90^{\circ} \\ \cline{1-6} \sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1 \\ \cline{1-6} \cos \theta & 1 & \dfrac{\sqrt{3}}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{1}{2} & 0 \\ \cline{1-6} \tan\theta & 0 & \dfrac{1}{\sqrt{3}} & 1 & \sqrt{3} & \textsf{Not D{e}fined}\end{array}}$