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If $\sin \theta =\dfrac{3}{5}$, find the values of other trigonometric ratios.

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Last updated date: 20th Jun 2024
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Answer
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Hint:Consider a right angled $\Delta ABC$, right angle at B. Use Pythagoras theorem then find the values of $\cos \theta $, $\tan \theta $ and their reciprocal $\csc \theta $, $\sec \theta $ and $\cot \theta $.

Complete step-by-step answer:
Let us consider a right angled triangle ABC. We know the Pythagoras theorem, also known as Pythagoras theorem; it is a fundamental relation in Euclidean geometry among the 3 sides of a right triangle.
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It states that the area of the squares whose sides is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.
$A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$
We have been given, $\sin \theta =\dfrac{3}{5}$.
In the triangle ABC, $\angle B=90{}^\circ $
And take $\angle C=\theta $
Here, $\sin \theta =$opposite side/hypotenuse = $\dfrac{AB}{BC}$
$\cos \theta =$Adjacent side/hypotenuse = $\dfrac{BC}{AC}$
Given,$\sin \theta =\dfrac{3}{5}$
AB=3 and AC=5
Using the Pythagoras theorem, $A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$
$\begin{align}
  & {{5}^{2}}={{3}^{2}}+B{{C}^{2}} \\
 & \Rightarrow B{{C}^{2}}={{5}^{2}}-{{3}^{2}} \\
 & \Rightarrow BC=\sqrt{25-9}=\sqrt{16}=4 \\
 & \therefore \cos \theta =\dfrac{4}{5} \\
\end{align}$
$\tan \theta =$Opposite side/adjacent side=$\dfrac{AB}{BC}=\dfrac{3}{4}$
$\cos ec\theta =\dfrac{1}{\sin \theta }=\dfrac{1}{\dfrac{3}{5}}=\dfrac{5}{3}$
$\begin{align}
  & \sec \theta =\dfrac{1}{\cos \theta }=\dfrac{1}{\dfrac{4}{5}}=\dfrac{5}{4} \\
 & \cot \theta =\dfrac{1}{\tan \theta }=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3} \\
 & \therefore \sin \theta =\dfrac{3}{5},\cos \theta =\dfrac{4}{5},\tan \theta =\dfrac{3}{4} \\
 & \cos ec\theta =\dfrac{5}{3},\sec \theta =\dfrac{5}{4},\cot \theta =\dfrac{4}{3} \\
\end{align}$

Note: There are three types of special right triangle, 30-60-90 triangle, 45-45-90 triangle and Pythagoras triple triangles.