Answer
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Hint:n order to determine exact values of all six trigonometric function of the angle whose cosine is given to us in the above question, first assume the a right angled triangle and express all the trigonometric ratios with respect to and in terms of the sides of that triangle and then find all the trigonometric ratios knowing the lengths of hypotenuse, altitude and base.
Complete step by step solution:
We are given a point $\cos x = \dfrac{3}{5}$. Now we assume a right angled triangle ABC and let $\angle BAC = x$
Now, $\angle ACB = {90^ \circ }$.
So, $\cos x = \dfrac{{Base}}{{{\text{Hypotenuse}}}} = \dfrac{3}{5}$.
So, we know the ratio of Base and Hypotenuse. Let ${\text{Base = 3x}}$ and ${\text{Hypotenuse = 5x}}$.
Now, calculating opposite side of the triangle using Pythagoras theorem,
${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Altitude} \right)^2}$
$ \Rightarrow {\left( {5x} \right)^2} = {\left( {3x} \right)^2} + {\left( {Altitude} \right)^2}$
$ \Rightarrow {\left( {Altitude} \right)^2} = 25{x^2} - 9{x^2}$
$ \Rightarrow {\left( {Altitude} \right)^2} = 16{x^2}$
$ \Rightarrow Altitude = 4x$
Therefore Calculating all the trigonometric ratios as:
\[\sin x = \dfrac{{{\text{Altitude}}}}{{{\text{Hypotenuse}}}} = \dfrac{4}{5}\]
\[\tan x = \dfrac{{{\text{Altitude}}}}{{{\text{Base}}}} = \dfrac{4}{3}\]
\[\cot x = \dfrac{{{\text{Base}}}}{{{\text{Altitude}}}} = \dfrac{3}{4}\]
\[\cos ecx = \dfrac{{{\text{Hypotenuse}}}}{{{\text{Altitude}}}} = \dfrac{5}{4}\]
\[\sec x = \dfrac{{{\text{Hypotenuse}}}}{{{\text{Base}}}} = \dfrac{5}{3}\]
So, these are the values of trigonometric ratios except cosine which was given beforehand in the question itself.
Note: Trigonometry is one of the significant branches throughout the entire existence of mathematics and has wide ranging applications in various fields of mathematics such as Geometry, Algebra and Calculus. One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer. Trigonometric ratios are the ratios of the sides of a triangle and thus the trigonometric ratios can be found by expressing the ratios in the terms of the sides of a triangle.
Complete step by step solution:
We are given a point $\cos x = \dfrac{3}{5}$. Now we assume a right angled triangle ABC and let $\angle BAC = x$
![seo images](https://www.vedantu.com/question-sets/f07704a3-785a-4281-93b9-b44c7c76c82d7297940235820231086.png)
Now, $\angle ACB = {90^ \circ }$.
So, $\cos x = \dfrac{{Base}}{{{\text{Hypotenuse}}}} = \dfrac{3}{5}$.
So, we know the ratio of Base and Hypotenuse. Let ${\text{Base = 3x}}$ and ${\text{Hypotenuse = 5x}}$.
Now, calculating opposite side of the triangle using Pythagoras theorem,
${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Altitude} \right)^2}$
$ \Rightarrow {\left( {5x} \right)^2} = {\left( {3x} \right)^2} + {\left( {Altitude} \right)^2}$
$ \Rightarrow {\left( {Altitude} \right)^2} = 25{x^2} - 9{x^2}$
$ \Rightarrow {\left( {Altitude} \right)^2} = 16{x^2}$
$ \Rightarrow Altitude = 4x$
Therefore Calculating all the trigonometric ratios as:
\[\sin x = \dfrac{{{\text{Altitude}}}}{{{\text{Hypotenuse}}}} = \dfrac{4}{5}\]
\[\tan x = \dfrac{{{\text{Altitude}}}}{{{\text{Base}}}} = \dfrac{4}{3}\]
\[\cot x = \dfrac{{{\text{Base}}}}{{{\text{Altitude}}}} = \dfrac{3}{4}\]
\[\cos ecx = \dfrac{{{\text{Hypotenuse}}}}{{{\text{Altitude}}}} = \dfrac{5}{4}\]
\[\sec x = \dfrac{{{\text{Hypotenuse}}}}{{{\text{Base}}}} = \dfrac{5}{3}\]
So, these are the values of trigonometric ratios except cosine which was given beforehand in the question itself.
Note: Trigonometry is one of the significant branches throughout the entire existence of mathematics and has wide ranging applications in various fields of mathematics such as Geometry, Algebra and Calculus. One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer. Trigonometric ratios are the ratios of the sides of a triangle and thus the trigonometric ratios can be found by expressing the ratios in the terms of the sides of a triangle.
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