Question

If we consider only the principle values of the inverse trigonometric functions then what will be the value of$\tan \left[ {{{\cos }^{ - 1}}\left( {\dfrac{1}{{5\sqrt 2 }}} \right) - {{\sin }^{ - 1}}\left( {\dfrac{4}{{\sqrt {17} }}} \right)} \right]$?
A. $\dfrac{3}{{29}}$
B. Cannot be determined
C. $- \dfrac{3}{{29}}$
D. $\dfrac{{29}}{3}$

Answer 1

The given problem revolves around the concepts of the equation of trigonometric ratios and its identities. To find the respective value of the given expression, we need to make the problem as easy. So, we will first assume any variables for inside terms particularly, then using the trigonometric identity i.e. ${\sin ^2}\theta + {\cos ^2}\theta = 1$(for both the terms) substituting the trigonometric ratio for compound angles i.e. $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$ the desired value can be obtained.

Complete step by step answer:
Since, we have given the expression as,
$\tan \left[ {{{\cos }^{ - 1}}\left( {\dfrac{1}{{5\sqrt 2 }}} \right) - {{\sin }^{ - 1}}\left( {\dfrac{4}{{\sqrt {17} }}} \right)} \right]$
To find the desired value we need to know the respective values in the bracket.Therefore, considering
${\cos ^{ - 1}}\left( {\dfrac{1}{{5\sqrt 2 }}} \right) = A$
And, ${\sin ^{ - 1}}\left( {\dfrac{4}{{\sqrt {17} }}} \right) = B$… ($1$)
which seems to be equal that,
$\cos A = \dfrac{1}{{5\sqrt 2 }} \\\ \Rightarrow A = {\cos ^{ - 1}}\left( {\dfrac{1}{{5\sqrt 2 }}} \right) \\\$
And

\Rightarrow B = {\sin ^{ - 1}}\left( {\dfrac{4}{{\sqrt {17} }}} \right) \\\ $$ As a result, finding the respective terms in sine and cosine using the trigonometric identity$${\sin ^2}\theta + {\cos ^2}\theta = 1$$, we get Considering $\cos A = \dfrac{1}{{5\sqrt 2 }}$, substituting it into the above identity, we get $${\sin ^2}{\rm A} = 1 - {\cos ^2}{\rm A} \\\ \Rightarrow {\sin ^2}{\rm A} = 1 - {\left( {\dfrac{1}{{5\sqrt 2 }}} \right)^2} \\\ $$ Solving the equation mathematically, we get $${\sin ^2}{\rm A} = 1 - \dfrac{1}{{50}} = \dfrac{{49}}{{50}} \\\ \Rightarrow \sin {\rm A} = \dfrac{7}{{5\sqrt 2 }} \\\ $$ … (i) Hence, by substituting this value in the identity$${\sin ^2}\theta + {\cos ^2}\theta = 1$$, we get $$\dfrac{{49}}{{50}} + {\cos ^2}{\rm A} = 1 \\\ \Rightarrow {\cos ^2}{\rm A} = 1 - \dfrac{{49}}{{50}} = \dfrac{1}{{50}} \\\ $$ Taking square roots, we get $$\cos {\rm A} = \dfrac{1}{{5\sqrt 2 }}$$ … (ii) From (i) and (ii) $\tan A = \dfrac{{\sin A}}{{\cos A}}$, we get $$\tan A = \dfrac{{\sin A}}{{\cos A}} = \dfrac{{\dfrac{7}{{5\sqrt 2 }}}}{{\dfrac{1}{{5\sqrt 2 }}}} \\\ \Rightarrow \tan A = 7 \\\ $$ … ($2$) Similarly, Considering $$\sin B = \dfrac{4}{{\sqrt {17} }}$$, substituting it into the above identity, we get $${\cos ^2}{\rm B} = 1 - {\sin ^2}{\rm B} \\\ \Rightarrow {\cos ^2}{\rm A} = 1 - {\left( {\dfrac{4}{{\sqrt {17} }}} \right)^2} \\\ $$ Solving the equation mathematically, we get $${\cos ^2}{\rm B} = 1 - \dfrac{{16}}{{17}} = \dfrac{1}{{17}} \\\ \Rightarrow \cos B = \dfrac{1}{{\sqrt {17} }} \\\ $$ … (iii) Hence, by substituting this value in the identity $${\sin ^2}\theta + {\cos ^2}\theta = 1$$, we get $${\sin ^2}{\rm B} + \dfrac{1}{{17}} = 1 \\\ \Rightarrow {\sin ^2}{\rm B} = 1 - \dfrac{1}{{17}} = \dfrac{{16}}{{17}} \\\ $$ Taking square roots, we get $$\sin {\rm B} = \dfrac{4}{{\sqrt {17} }}$$ … (iv) From (iii) and (iv) $$\tan B = \dfrac{{\sin B}}{{\cos B}}$$, we get $$\tan B = \dfrac{{\sin B}}{{\cos B}} = \dfrac{{\dfrac{4}{{\sqrt {17} }}}}{{\dfrac{1}{{\sqrt {17} }}}} \\\ \Rightarrow \tan B = 4 $$ … ($3$) Now, considering the given expression i.e. $\tan \left[ {{{\cos }^{ - 1}}\left( {\dfrac{1}{{5\sqrt 2 }}} \right) - {{\sin }^{ - 1}}\left( {\dfrac{4}{{\sqrt {17} }}} \right)} \right]$ Hence, from ($1$), it is assumed that As a result, the expression becomes $\tan \left[ {{{\cos }^{ - 1}}\left( {\dfrac{1}{{5\sqrt 2 }}} \right) - {{\sin }^{ - 1}}\left( {\dfrac{4}{{\sqrt {17} }}} \right)} \right] = \tan \left[ {A - B} \right]$ We know that, $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$ As a result, from ($2$) and ($3$), we get $\tan \left( {A - B} \right) = \dfrac{{7 - 4}}{{1 + 7 \times 4}} \\\ \Rightarrow \tan \left( {A - B} \right) = \dfrac{3}{{29}} \\\ $ Re-substituting the values, we get $\tan \left[ {{{\cos }^{ - 1}}\left( {\dfrac{1}{{5\sqrt 2 }}} \right) - {{\sin }^{ - 1}}\left( {\dfrac{4}{{\sqrt {17} }}} \right)} \right] = \tan \left( {A - B} \right) \\\ \therefore \tan \left[ {{{\cos }^{ - 1}}\left( {\dfrac{1}{{5\sqrt 2 }}} \right) - {{\sin }^{ - 1}}\left( {\dfrac{4}{{\sqrt {17} }}} \right)} \right] = \dfrac{3}{{29}}$ **Hence, option A is correct.** **Note:** One must know all the trigonometric ratios, identities for the ease of the problem such as $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$, $${\sin ^2}\theta + {\cos ^2}\theta = 1$$, etc. To calculate the final value for tan parameter, convert the respective sine and cosine terms by the trigonometric parameter$$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$$ so as to be sure of our final answer.