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

If cosA=941\cos A = \dfrac{9}{{41}}, find
tanA\tan A and cscA\csc A.

Explanation

Solution

Here, we need to find the value of tanA\tan A and cscA\csc A. We will use the trigonometric identities and ratios to solve the question. We will use the formula sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 to find the value of sine of AA. Using the value of sine of AA, we will find the value of cosecant of AA. Using the value of sine and cosine of AA, we can find the value of the tangent of AA.

Formula Used: The sum of squares of the sine and cosine of an angle is equal to 1, that is sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1.
The tangent of an angle θ\theta is the ratio of the sine and cosine of the angle θ\theta , that is tanθ=sinθcosθ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}.
The cosecant of an angle θ\theta is the reciprocal of the sine of the angle θ\theta , that is cscθ=1sinθ\csc \theta = \dfrac{1}{{\sin \theta }}.

Complete step-by-step answer:
We can find the value of tanA\tan A and cscA\csc A if we have the values of sinA\sin A and cosA\cos A.
First, we will find the value of sine of angle of AA.
As we know sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1.
Therefore, we can write sin2A+cos2A=1{\sin ^2}A + {\cos ^2}A = 1.
Substituting the given value cosA=941\cos A = \dfrac{9}{{41}}, we get
sin2A+(941)2=1\Rightarrow {\sin ^2}A + {\left( {\dfrac{9}{{41}}} \right)^2} = 1
Simplifying the expression, we get
sin2A+811681=1\Rightarrow {\sin ^2}A + \dfrac{{81}}{{1681}} = 1
Subtracting 811681\dfrac{{81}}{{1681}} from both sides of the equation, we get
sin2A+811681811681=1811681 sin2A=1681811681 sin2A=16001681\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
sinA=16001681 sinA=4041\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 cscA\csc A.
The cosecant of an angle θ\theta is the reciprocal of the sine of the angle θ\theta , that is cscθ=1sinθ\csc \theta = \dfrac{1}{{\sin \theta }}.
Therefore, we get
cscA=1sinA\csc A = \dfrac{1}{{\sin A}}
Substituting the value sinA=4041\sin A = \dfrac{{40}}{{41}}, we get
cscA=14041\Rightarrow \csc A = \dfrac{1}{{\dfrac{{40}}{{41}}}}
Simplifying the expression, we get
cscA=4140\Rightarrow \csc A = \dfrac{{41}}{{40}}
Next, we can find the value of tanA\tan A.
The tangent of an angle
θ\theta is the ratio of the sine and cosine of the angle θ\theta , that is tanθ=sinθcosθ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}.
Therefore, we get
tanA=sinAcosA\tan A = \dfrac{{\sin A}}{{\cos A}}
Substituting the value sinA=4041\sin A = \dfrac{{40}}{{41}} and cosA=941\cos A = \dfrac{9}{{41}}, we get
tanA=4041941\Rightarrow \tan A = \dfrac{{\dfrac{{40}}{{41}}}}{{\dfrac{9}{{41}}}}
Simplifying the expression, we get
tanA=409\therefore \tan A=\dfrac{40}{9}
Therefore, the value of tanA\tan A and
cscA\csc A is 409\dfrac{{40}}{9} and 4140\dfrac{{41}}{{40}} respectively.

Note: We can also find the values of tanA\tan A and cscA\csc A using the definitions of the trigonometric ratios.
We know that cosine of an angle θ\theta in a right angled triangle is given by cosθ=BaseHypotenuse\cos \theta = \dfrac{{{\rm{Base}}}}{{{\rm{Hypotenuse}}}}.
Since cosA=941\cos A = \dfrac{9}{{41}}, assume that base =9x = 9x and hypotenuse =41x = 41x.
Using the Pythagoras’s theorem, we get
(Hypotenuse)2=(Base)2+(Perpendicular)2 (41x)2=(9x)2+(Perpendicular)2\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
1681x2=81x2+(Perpendicular)2 (Perpendicular)2=1681x281x2 (Perpendicular)2=1600x2\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
Perpendicular=40x\Rightarrow {\rm{Perpendicular}} = 40x
Now, we know that the cosecant of an angle θ\theta in a right angled triangle is given by cscθ=HypotenusePerpendicular\csc \theta = \dfrac{{{\rm{Hypotenuse}}}}{{{\rm{Perpendicular}}}}.
Substituting the values of the hypotenuse and perpendicular, we get
cscA=41x40x=4140\Rightarrow \csc A = \dfrac{{41x}}{{40x}} = \dfrac{{41}}{{40}}
Also, we know that the tangent of an angle θ\theta in a right angled triangle is given by tanθ=PerpendicularBase\tan \theta = \dfrac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}}.
Substituting the values of the hypotenuse and perpendicular, we get
tanA=40x9x=409\Rightarrow \tan A = \dfrac{{40x}}{{9x}} = \dfrac{{40}}{9}
Therefore, the value of tanA\tan A and cscA\csc A is 409\dfrac{{40}}{9} and 4140\dfrac{{41}}{{40}} respectively.