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 use the table which gives us the value of trigonometric functions at different angles.
With the help of the table we find the values of \[\theta \] and \[\phi \]. Substitute the values of obtained angles in the numerator and denominator and write the values of trigonometric functions using the table.
* Table for trigonometric functions like sine, cosine and tan at angles \[{0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }\] is

ANGLEFUNCTION	\[{0^ \circ }\]	\[{30^ \circ }\]	\[{45^ \circ }\]	\[{60^ \circ }\]	\[{90^ \circ }\]
Sin	0	\[\dfrac{1}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{{\sqrt 3 }}{2}\]	1
Cos	1	\[\dfrac{{\sqrt 3 }}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{1}{2}\]	0
Tan	0	\[\dfrac{1}{{\sqrt 3 }}\]	1	\[\sqrt 3 \]	Not defined

Complete step-by-step answer:
We are given that \[\sin \theta = \dfrac{{\sqrt 3 }}{2}\]
So, here the angle is \[\theta \]
We know that the value of sine function is \[\dfrac{{\sqrt 3 }}{2}\]when the angle is \[{60^ \circ }\]
Therefore, we can write value of \[\dfrac{{\sqrt 3 }}{2} = \sin {60^ \circ }\]
\[ \Rightarrow \sin \theta = \sin {60^ \circ }\]
We know the value of trigonometric function will be equal if the angles will be of equal measure
\[ \Rightarrow \theta = {60^ \circ }\] … (1)
Also,\[\cos \phi = \dfrac{1}{{\sqrt 2 }}\]
So, here the angle is \[\phi \]
We know that the value of cosine function is \[\dfrac{1}{{\sqrt 2 }}\]when the angle is \[{45^ \circ }\]
Therefore, we can write value of \[\dfrac{1}{{\sqrt 2 }} = \cos {45^ \circ }\]
\[ \Rightarrow \cos \phi = \cos {45^ \circ }\]
We know the value of trigonometric function will be equal if the angles will be of equal measure
\[ \Rightarrow \phi = {45^ \circ }\] … (2)
Now we have to find the value of \[\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}\]
Substitute the values of angle \[\theta \]from equation (1) and angle \[\phi \]from equation (2) in numerator and denominator of \[\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}\]
\[ \Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\tan {{60}^ \circ } - \tan {{45}^ \circ }}}{{1 + \tan {{60}^ \circ }\tan {{45}^ \circ }}}\] … (3)
Now we find the values of \[\tan {60^ \circ }\]and\[\tan {45^ \circ }\]from the table.
\[\tan {60^ \circ } = \sqrt 3 \]and\[\tan {45^ \circ } = 1\]
Substitute the values in numerator and denominator of equation (3)
\[ \Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 \times 1}}\]
\[ \Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\]
Rationalize the term in RHS by multiplying both numerator and denominator by \[\sqrt 3 - 1\]
\[ \Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}\]
Multiply the numerator with numerator and denominator with denominator
\[ \Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}\]
\[ \Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}\]
Use the property\[{(a - b)^2} = {a^2} + {b^2} - 2ab\]in numerator and the property \[(a + b)(a - b) = {a^2} - {b^2}\]in denominator of RHS
\[ \Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\left( {{{(\sqrt 3 )}^2} + {{(1)}^2} - 2 \times (\sqrt 3 \times 1)} \right)}}{{{{(\sqrt 3 )}^2} - {{(1)}^2}}}\]
\[ \Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{3 + 1 - 2\sqrt 3 }}{{3 - 1}}\]
\[ \Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{4 - 2\sqrt 3 }}{2}\]
Take 2 common from numerator in RHS of the equation
\[ \Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{2(2 - \sqrt 3 )}}{2}\]
Cancel the same factors from numerator and denominator.
\[ \Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = 2 - \sqrt 3 \]

So, the value of \[\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}\] is \[2 - \sqrt 3 \].

Note: Students might try to use the property of \[\tan A - \tan B = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\] to write \[\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}\]. But then we will have the term equal to \[\tan (\theta - \phi ) = \tan {15^ \circ }\]. Since, we don’t know the value of \[\tan {15^ \circ }\] we avoid this process. Also, many students leave the answer with an irrational number i.e. under root value in the denominator, keep in mind the answer should always be rationalized and in lowest form.