Question

If \[f\left( \theta \right) = \dfrac{{1 - \sin 2\theta + \cos 2\theta }}{{2\cos 2\theta }}\] then the value of \[f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right)\]
A) \[\dfrac{1}{2}\]
B) \[\dfrac{3}{4}\]
C) \[\dfrac{1}{4}\]
D) 1

Answer 1

Hint:
We will solve this question by splitting the numerator in two parts and then applying various trigonometric formulas and half angle formulas of \[\sin \theta \] and \[\cos \theta \] respectively.

Formulas used:
We will use following formulas:
1) \[\sin 2\theta = 2\sin \theta \cdot \cos \theta \]
2) \[\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \]
3) \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
4) \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
5) \[\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)\]
6) \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
7) \[\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a \cdot \tan b}}\]

Complete step by step solution:
According to the question,
 \[f\left( \theta \right) = \dfrac{{1 - \sin 2\theta + \cos 2\theta }}{{2\cos 2\theta }}\]
Now, splitting the numerator of R.H.S. in two parts, we get
\[ \Rightarrow f\left( \theta \right) = \dfrac{{1 - \sin 2\theta }}{{2\cos 2\theta }} + \dfrac{{\cos 2\theta }}{{2\cos 2\theta }}\]
We will now use the trigonometric formula \[\sin 2\theta = 2\sin \theta \cdot \cos \theta \] and \[\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \] to expand the terms of the above equation.
 Therefore, we get
\[ \Rightarrow f\left( \theta \right) = \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta - 2\sin \theta \cdot \cos \theta }}{{2\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)}} + \dfrac{{\cos 2\theta }}{{2\cos 2\theta }}\]
As we can notice, the numerator is in the form of \[{a^2} + {b^2} - 2ab\]and the denominator is in the form of \[\left( {{a^2} - {b^2}} \right)\], hence, applying the formula \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] and \[\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)\]in the numerator and denominator respectively, we get
\[ \Rightarrow f\left( \theta \right) = \dfrac{{{{\left( {\cos \theta - \sin \theta } \right)}^2}}}{{2\left( {\cos \theta - \sin \theta } \right)\left( {\cos \theta + \sin \theta } \right)}} + \dfrac{{\cos 2\theta }}{{2\cos 2\theta }}\]
Now, eliminating the same parts of numerator and denominator in RHS, we get
\[ \Rightarrow f\left( \theta \right) = \dfrac{{\left( {\cos \theta - \sin \theta } \right)}}{{2\left( {\cos \theta + \sin \theta } \right)}} + \dfrac{1}{2}\]
Taking \[\dfrac{1}{2}\] common, we get
\[ \Rightarrow f\left( \theta \right) = \dfrac{1}{2}\left[ {\dfrac{{\left( {\cos \theta - \sin \theta } \right)}}{{\left( {\cos \theta + \sin \theta } \right)}} + 1} \right]\]
Dividing numerator and denominator by \[\cos \theta \], we get
\[ \Rightarrow f\left( \theta \right) = \dfrac{1}{2}\left[ {\dfrac{{\left( {\dfrac{{\cos \theta }}{{\cos \theta }} - \dfrac{{\sin \theta }}{{\cos \theta }}} \right)}}{{\left( {\dfrac{{\cos \theta }}{{\cos \theta }} + \dfrac{{\sin \theta }}{{\cos \theta }}} \right)}} + 1} \right]\]
Now, using \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] in the above equation, we get
\[ \Rightarrow f\left( \theta \right) = \dfrac{1}{2}\left[ {\dfrac{{\left( {1 - \tan \theta } \right)}}{{\left( {1 + \tan \theta } \right)}} + 1} \right]\]
Substituting \[1 = \tan 45^\circ \] in the above equation , we get
\[f\left( \theta \right) = \dfrac{1}{2}\left[ {\dfrac{{\left( {\tan 45^\circ - \tan \theta } \right)}}{{\left( {1 + \tan 45^\circ \cdot \tan \theta } \right)}} + 1} \right]\]
Using the formula \[\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a \cdot \tan b}}\], we can simplify above equation as

\[ \Rightarrow f\left( \theta \right) = \dfrac{1}{2}\left[ {\tan \left( {45^\circ - \theta } \right) + 1} \right]\]……………………………\[\left( 1 \right)\]
Now we will find \[f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right)\] by substituting the value of \[\theta \] as \[11^\circ \] and \[34^\circ \] the equation\[\left( 1 \right)\].
\[f\left( {11^\circ } \right) = \dfrac{1}{2}\left[ {\tan \left( {45^\circ - 11^\circ } \right) + 1} \right]\]
and
\[f\left( {34^\circ } \right) = \dfrac{1}{2}\left[ {\tan \left( {45^\circ - 34^\circ } \right) + 1} \right]\]
Multiplying the two equations, we get
\[f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{2}\left[ {\tan \left( {45^\circ - 11^\circ } \right) + 1} \right] \cdot \dfrac{1}{2}\left[ {\tan \left( {45^\circ - 34^\circ } \right) + 1} \right]\]
\[ \Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{4}\left[ {\tan \left( {34^\circ } \right) + 1} \right] \cdot \left[ {\tan \left( {11^\circ } \right) + 1} \right]\]
Now, multiplying both the brackets on RHS, we get
\[ \Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{4}\left[ {\tan \left( {34^\circ } \right) \cdot \tan \left( {11^\circ } \right) + \tan \left( {34^\circ } \right) + \tan \left( {11^\circ } \right) + 1} \right]\]…………………………\[\left( 2 \right)\]
Now as we know, \[\tan \left( {45^\circ } \right) = \tan \left( {34^\circ + 11^\circ } \right)\], so using the identity, we get
\[1 = \dfrac{{\tan 34^\circ + \tan 11^\circ }}{{1 - \tan 34^\circ \cdot \tan 11^\circ }}\]
\[ \Rightarrow 1 = \dfrac{{\tan 34^\circ + \tan 11^\circ }}{{1 - \tan 34^\circ \cdot \tan 11^\circ }}\]
\[ \Rightarrow 1 - \tan 34^\circ \cdot \tan 11^\circ = \tan 34^\circ + \tan 11^\circ \]
Now substituting the value of \[\tan 34^\circ + \tan 11^\circ \] in equation number \[\left( 2 \right)\], we get
\[ \Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{4}\left[ {\tan \left( {34^\circ } \right) \cdot \tan \left( {11^\circ } \right) + 1 - \tan \left( {34^\circ } \right) \cdot \tan \left( {11^\circ } \right) + 1} \right]\]
Now eliminating the same constants, we get,
\[ \Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{4}\left[ {1 + 1} \right]\]
\[ \Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{2}{4}\]
Dividing numerator and denominator by \[2\], we get
\[ \Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{2}\]
Hence, the required value of \[f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right)\] is \[\dfrac{1}{2}\].

Hence, option A is the correct option.

Note:
If we are aware of all the basic and trigonometric formulas, then this question can be solved easily. This question can be a bit tricky at the beginning because it’s quite tough to recognize that from where we should begin solving this question. Also, keeping in mind the fact that the numerator can be rewritten using the formula \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] and other basic formulas. We should be aware of calculation mistakes in these types of questions as it is possible that we would use ‘plus’ sign instead of using a ‘minus’ and vice-versa in various formulas. Also, the fact that \[\tan \left( {45^\circ } \right) = \tan \left( {34^\circ + 11^\circ } \right)\] helped us to further solve this question. If this basic addition would have not striked the mind then, this question could be a tough one to solve. Hence, all these facts make this question quite necessary.