Question

If \[\cos A = \dfrac{9}{{41}}\], find
\[\tan A\] and \[\csc A\].

Answer 1

Hint: Here, we need to find the value of \[\tan A\] and \[\csc A\]. We will use the trigonometric identities and ratios to solve the question. We will use the formula \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] to find the value of sine of \[A\]. Using the value of sine of \[A\], we will find the value of cosecant of \[A\]. Using the value of sine and cosine of \[A\], we can find the value of the tangent of \[A\].

Formula Used: The sum of squares of the sine and cosine of an angle is equal to 1, that is \[{\sin ^2}\theta + {\cos ^2}\theta = 1\].
The tangent of an angle \[\theta \] is the ratio of the sine and cosine of the angle \[\theta \], that is \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\].
The cosecant of an angle \[\theta \] is the reciprocal of the sine of the angle \[\theta \], that is \[\csc \theta = \dfrac{1}{{\sin \theta }}\].

Complete step-by-step answer:
We can find the value of \[\tan A\] and \[\csc A\] if we have the values of \[\sin A\] and \[\cos A\].
First, we will find the value of sine of angle of \[A\].
As we know \[{\sin ^2}\theta + {\cos ^2}\theta = 1\].
Therefore, we can write \[{\sin ^2}A + {\cos ^2}A = 1\].
Substituting the given value \[\cos A = \dfrac{9}{{41}}\], we get
\[ \Rightarrow {\sin ^2}A + {\left( {\dfrac{9}{{41}}} \right)^2} = 1\]
Simplifying the expression, we get
\[ \Rightarrow {\sin ^2}A + \dfrac{{81}}{{1681}} = 1\]
Subtracting \[\dfrac{{81}}{{1681}}\] from both sides of the equation, we get
\[\begin{array}{l} \Rightarrow {\sin ^2}A + \dfrac{{81}}{{1681}} - \dfrac{{81}}{{1681}} = 1 - \dfrac{{81}}{{1681}}\\ \Rightarrow {\sin ^2}A = \dfrac{{1681 - 81}}{{1681}}\\ \Rightarrow {\sin ^2}A = \dfrac{{1600}}{{1681}}\end{array}\]
Taking the square root on both the sides, we get
\[\begin{array}{l} \Rightarrow \sin A = \sqrt {\dfrac{{1600}}{{1681}}} \\ \Rightarrow \sin A = \dfrac{{40}}{{41}}\end{array}\]
Now, we can find the value of \[\csc A\].
The cosecant of an angle \[\theta \] is the reciprocal of the sine of the angle \[\theta \], that is \[\csc \theta = \dfrac{1}{{\sin \theta }}\].
Therefore, we get
\[\csc A = \dfrac{1}{{\sin A}}\]
Substituting the value \[\sin A = \dfrac{{40}}{{41}}\], we get
\[ \Rightarrow \csc A = \dfrac{1}{{\dfrac{{40}}{{41}}}}\]
Simplifying the expression, we get
\[ \Rightarrow \csc A = \dfrac{{41}}{{40}}\]
Next, we can find the value of \[\tan A\].
The tangent of an angle
\[\theta \] is the ratio of the sine and cosine of the angle \[\theta \], that is \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\].
Therefore, we get
\[\tan A = \dfrac{{\sin A}}{{\cos A}}\]
Substituting the value \[\sin A = \dfrac{{40}}{{41}}\] and \[\cos A = \dfrac{9}{{41}}\], we get
\[ \Rightarrow \tan A = \dfrac{{\dfrac{{40}}{{41}}}}{{\dfrac{9}{{41}}}}\]
Simplifying the expression, we get
\[\therefore \tan A=\dfrac{40}{9}\]
Therefore, the value of \[\tan A\] and
\[\csc A\] is \[\dfrac{{40}}{9}\] and \[\dfrac{{41}}{{40}}\] respectively.

Note: We can also find the values of \[\tan A\] and \[\csc A\] using the definitions of the trigonometric ratios.
We know that cosine of an angle \[\theta \] in a right angled triangle is given by \[\cos \theta = \dfrac{{{\rm{Base}}}}{{{\rm{Hypotenuse}}}}\].
Since \[\cos A = \dfrac{9}{{41}}\], assume that base \[ = 9x\] and hypotenuse \[ = 41x\].
Using the Pythagoras’s theorem, we get
\[\begin{array}{l}{\rm{(Hypotenuse)}}{{\rm{}}^2} = {\rm{(Base)}}{{\rm{}}^2} + {\rm{(Perpendicular)}}{{\rm{}}^2}\\ \Rightarrow {\left( {41x} \right)^2} = {\left( {9x} \right)^2} + {\rm{(Perpendicular)}}{{\rm{}}^2}\end{array}\]
Solving the above equation to find the perpendicular, we get
\[\begin{array}{l} \Rightarrow 1681{x^2} = 81{x^2} + {\rm{(Perpendicular)}}{{\rm{}}^2}\\ \Rightarrow {\rm{(Perpendicular)}}{{\rm{}}^2} = 1681{x^2} - 81{x^2}\\ \Rightarrow {\rm{(Perpendicular)}}{{\rm{}}^2} = 1600{x^2}\end{array}\]
Taking square root of both sides, we get
\[ \Rightarrow {\rm{Perpendicular}} = 40x\]
Now, we know that the cosecant of an angle \[\theta \] in a right angled triangle is given by \[\csc \theta = \dfrac{{{\rm{Hypotenuse}}}}{{{\rm{Perpendicular}}}}\].
Substituting the values of the hypotenuse and perpendicular, we get
\[ \Rightarrow \csc A = \dfrac{{41x}}{{40x}} = \dfrac{{41}}{{40}}\]
Also, we know that the tangent of an angle \[\theta \] in a right angled triangle is given by \[\tan \theta = \dfrac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}}\].
Substituting the values of the hypotenuse and perpendicular, we get
\[ \Rightarrow \tan A = \dfrac{{40x}}{{9x}} = \dfrac{{40}}{9}\]
Therefore, the value of \[\tan A\] and \[\csc A\] is \[\dfrac{{40}}{9}\] and \[\dfrac{{41}}{{40}}\] respectively.