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**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.

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