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

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 $\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$.