Question

If \[\sin 4A - \cos 2A = \cos 4A - \sin 2A\] (where, \[0 < A < \dfrac{\pi }{4}\] ) then the value of \[\tan 4A\] is
A) \[1\]
B) \[\dfrac{1}{{\sqrt 3 }}\]
C) \[\sqrt 3 \]
D) \[\dfrac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}}\]
E) \[\dfrac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}}\]

Answer 1

Hint: Here in this question, we have to find the exact value of the given equation of trigonometric function. For this first we have to simplify the given equation by using a Sum to Product Formula of trigonometry i.e., \[\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)\] and \[\cos x + \cos y = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)\] then by the definition and standard angles values of trigonometric ratios we get the required value.

Complete step by step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function.
Consider the given equation:
\[\sin 4A - \cos 2A = \cos 4A - \sin 2A\]
On rearranging the given equation, we have
\[ \Rightarrow \,\,\,\,\sin 4A + \sin 2A = \cos 4A + \cos 2A\] --------(1)
Now, apply the sum to product formula of trigonometry i.e., \[\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)\] In LHS and
\[\cos x + \cos y = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)\] in RHS, then
Here, \[x = 4A\] and \[y = 2A\]
On substituting the \[x\] and \[y\] values in formula, then equation (1) becomes
\[ \Rightarrow \,\,\,\,2\sin \left( {\dfrac{{4A + 2A}}{2}} \right)\cos \left( {\dfrac{{4A - 2A}}{2}} \right) = 2\cos \left( {\dfrac{{4A + 2A}}{2}} \right)\cos \left( {\dfrac{{4A - 2A}}{2}} \right)\]
\[ \Rightarrow \,\,\,\,2\sin \left( {\dfrac{{6A}}{2}} \right)\cos \left( {\dfrac{{2A}}{2}} \right) = 2\cos \left( {\dfrac{{6A}}{2}} \right)\cos \left( {\dfrac{{2A}}{2}} \right)\]
On simplification, we get
\[ \Rightarrow \,\,\,\,2\sin \left( {3A} \right)\cos \left( A \right) = 2\cos \left( {3A} \right)\cos \left( A \right)\]
Divide both side by \[\cos \left( A \right)\], then we have
\[ \Rightarrow \,\,\,\,2\sin \left( {3A} \right) = 2\cos \left( {3A} \right)\]
Divide both side by \[\cos \left( {3A} \right)\], then
\[ \Rightarrow \,\,\,\,\dfrac{{2\sin \left( {3A} \right)}}{{2\operatorname{Cos} \left( {3A} \right)}} = 1\]
On simplification, we get
\[ \Rightarrow \,\,\,\,\dfrac{{\sin \left( {3A} \right)}}{{\operatorname{Cos} \left( {3A} \right)}} = 1\] -----(2)
By the definition of trigonometric ratio, tangent is the ratio between sine and cosine i.e., \[\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta \], then equation (2) becomes
\[ \Rightarrow \,\,\,\,\tan 3A = 1\] -----(3)
Given the angle \[A\] ranges from \[0 < A < \dfrac{\pi }{4}\], then
\[3A = \dfrac{\pi }{4}\]
Both side divide by 3, then
\[A = \dfrac{\pi }{{12}}\]
Now find the value of \[\tan 4A\], then
\[ \Rightarrow \,\,\,\tan 4A\]
Substitute the value of \[A\], then
\[ \Rightarrow \,\,\,\tan 4\left( {\dfrac{\pi }{{12}}} \right)\]
On simplification, we get
\[ \Rightarrow \,\,\,\tan \left( {\dfrac{\pi }{3}} \right)\]
By the table of angle of standard trigonometric ratios the value of \[\tan {60^ \circ } = \tan \left( {\dfrac{\pi }{3}} \right) = \sqrt 3 \].
\[\therefore \,\,\,\tan \left( {\dfrac{\pi }{3}} \right) = \sqrt 3 \]
Hence, the required value is \[\sqrt 3 \].

Note:
When solving the trigonometry-based questions, we have to know the definitions and table of standard angles of all six trigonometric ratio’s sine, cosine, tangent, secant, cosecant and cotangent. Remember the standard formulas like trigonometric identities, double and half angles, sum and difference identity etc..