Question

If \[\sin \theta = \dfrac{{\sqrt 3 }}{2}\] and \[\cos \phi = \dfrac{1}{{\sqrt 2 }}\]. Find the value of \[\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}\].

Answer 1

Hint: Here, we will use the value \[\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}\] in the right side of the equation \[\sin \theta = \dfrac{{\sqrt 3 }}{2}\] and the value \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\] in the right side of the equation \[\cos\phi = \dfrac{1}{{\sqrt 2 }}\] to find the value of \[\theta \] and \[\phi \]. Then we will substitute these values in the equation, \[\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}\] and then simplify to find the required value.

Complete step by step answer:

We are given that
\[\sin \theta = \dfrac{{\sqrt 3 }}{2}{\text{ ......eq.(1)}}\]
 \[\cos \phi = \dfrac{1}{{\sqrt 2 }}{\text{ .......eq.(2)}}\].
Using the value \[\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}\] in the right side of the equation (1), we get
\[ \Rightarrow \sin \theta = \sin 60^\circ \]
Applying the \[{\sin ^{ - 1}}\] in the above equation, we get
\[ \Rightarrow {\sin ^{ - 1}}\sin \theta = {\sin ^{ - 1}}\sin 60^\circ \]
Using the trigonometric property, \[{\sin ^{ - 1}}\sin a = a\] in the above equation, we get
\[ \Rightarrow \theta = 60^\circ \]
Using the value \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\] in the right side of the equation (2), we get
\[ \Rightarrow \cos \phi = \cos 45^\circ \]
Applying the \[{\cos ^{ - 1}}\] in the above equation, we get
\[ \Rightarrow {\cos ^{ - 1}}\cos \phi = {\cos ^{ - 1}}\cos 45^\circ \]
Using the trigonometric property, \[{\cos ^{ - 1}}\cos a = a\] in the above equation, we get
\[ \Rightarrow \phi = 45^\circ \]
Substituting the value of \[\theta \] and \[\phi \] in the equation, \[\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}\], we get
\[ \Rightarrow \dfrac{{\tan 60^\circ - \tan 45^\circ }}{{1 + \tan 60^\circ \cdot \tan 45^\circ }}\]
Using the trigonometric value, \[\tan 60^\circ = \sqrt 3 \] and \[\tan 45^\circ = 1\] in the above equation and then simplify, we get
\[
   \Rightarrow \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 \cdot 1}} \\
   \Rightarrow \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }} \\
   \Rightarrow \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \\
 \]
Multiplying the numerator and denominator by \[\sqrt 3 - 1\] in the above equation, we get
\[
   \Rightarrow \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \cdot \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}} \\
   \Rightarrow \dfrac{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt {3 - 1} } \right)}} \\
   \Rightarrow \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt {3 - 1} } \right)}} \\
 \]
Using the value, \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\] in numerator and \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\] in denominator in the above equation, we get
\[
   \Rightarrow \dfrac{{{{\left( {\sqrt 3 } \right)}^2} - 2 \cdot 1 \cdot \sqrt 2 + {1^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {1^2}}} \\
   \Rightarrow \dfrac{{3 - 2\sqrt 2 + 1}}{{3 - 1}} \\
   \Rightarrow \dfrac{{4 - 2\sqrt 2 }}{2} \\
   \Rightarrow 2 - \sqrt 2 \\
 \]
Thus, the required value is \[2 - \sqrt 2 \].

Note: In solving these types of questions, the key concept is to have a good understanding of the basic trigonometric values and learn how to use the values from trigonometric tables. Students should have a grasp of trigonometric values, for simplifying the given equation. Remember that \[\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}\] and \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\].