Question

How do you evaluate \[\tan \left( {\arcsin \left( {\dfrac{3}{4}} \right)} \right)\]?

Answer 1

Hint: We use the concept that arc means inverse of the function. We are given the value of sine of an angle. Use the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\] to find the value of cosine of the same angle. Divide sine of the angle by cosine of the angle to calculate tangent of the angle.
* Arc sine of a function is defined as the inverse sine of the function. When \[\sin y = x\], then we can write \[\arcsin x = {\sin ^{ - 1}}x = y\].

Complete step-by-step answer:
Let us assume the angle as ‘x’
We are given that \[\arcsin \left( {\dfrac{3}{4}} \right)\]
Since we know that \[\arcsin x = {\sin ^{ - 1}}x = y\]
Then we can write \[si{n^{ - 1}}\left( {\dfrac{3}{4}} \right) = y\]
Take sine on both sides of the equation
\[ \Rightarrow \sin \left\{ {si{n^{ - 1}}\left( {\dfrac{3}{4}} \right)} \right\} = \sin y\]
Cancel sine and sine inverse on left side of the equation
\[ \Rightarrow \dfrac{3}{4} = \sin y\] … (1)
Now we know the identity \[{\sin ^2}x + {\cos ^2}x = 1\]
We can calculate the value of \[\cos y\]using the value from equation (1)
\[ \Rightarrow {\sin ^2}y + {\cos ^2}y = 1\]
Substitute the value of \[\sin y\]from equation (1)
\[ \Rightarrow {\left( {\dfrac{3}{4}} \right)^2} + {\cos ^2}y = 1\]
Square the terms in fraction
\[ \Rightarrow \dfrac{9}{{16}} + {\cos ^2}y = 1\]
Shift all constants to right hand side of the equation
\[ \Rightarrow {\cos ^2}y = 1 - \dfrac{9}{{16}}\]
Take LCM in right hand side of the equation
\[ \Rightarrow {\cos ^2}y = \dfrac{{16 - 9}}{{16}}\]
\[ \Rightarrow {\cos ^2}y = \dfrac{7}{{16}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{{\cos }^2}y} = \sqrt {\dfrac{7}{{16}}} \]
Write \[16 = {4^2}\]in the denominator
\[ \Rightarrow \sqrt {{{\cos }^2}y} = \sqrt {\dfrac{7}{{{4^2}}}} \]
Cancel square root by square power on both sides of the equation
\[ \Rightarrow \cos y = \dfrac{{\sqrt 7 }}{4}\] … (2)
Now we know that \[\tan y = \dfrac{{\sin y}}{{\cos y}}\]
Substitute the required values from equation (1) and (2)
\[ \Rightarrow \tan y = \dfrac{{\dfrac{3}{4}}}{{\dfrac{{\sqrt 7 }}{4}}}\]
Write fraction in simpler form
\[ \Rightarrow \tan y = \dfrac{3}{4} \times \dfrac{4}{{\sqrt 7 }}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow \tan y = \dfrac{3}{{\sqrt 7 }}\]
Now since we have irrational number in the denominator, we rationalize the fraction by multiplying both numerator and denominator by\[\sqrt 7 \]
\[ \Rightarrow \tan y = \dfrac{3}{{\sqrt 7 }} \times \dfrac{{\sqrt 7 }}{{\sqrt 7 }}\]
\[ \Rightarrow \tan y = \dfrac{{3\sqrt 7 }}{7}\]

\[\therefore \]The value of \[\tan \left( {\arcsin \left( {\dfrac{3}{4}} \right)} \right)\] is \[\dfrac{{3\sqrt 7 }}{7}\]

Note:
Many students make mistake of calculating the value of sine taking the angle as \[\dfrac{3}{4}\] because they think first we calculate sine of the angle and then apply arc function to it and then apply tangent to the obtained value which is wrong. Keep in mind arc sine indicates the inverse function of sine.