Question

If $ 2\tan {10^0} + \tan {50^0} = 2x,\,\tan {20^0} + \tan {50^0} = 2y,\,2\tan {10^0} + \tan {70^0} = 2w\,\,and\,\,\,\tan {20^0} + \tan {70^0} = 2z $ then which of the following is/are true. Solve given equations with each other to derive relationships between variables x, y, z and w to discuss given options.
(A) $ z > w > y > x $
(B) $ w = x + y $
(C) $ 2y = z $
(D) $ z + x = w + y $

Answer 1

Hint: In given problems four equations are given first we see common terms in equations and then group them together and solve them using method of elimination and finally obtaining relation between variable x, y, z and w to get required correct options.
 $ 2.\sin A.\cos B = \sin (A + B) + \sin (A - B),\,\,2\cos A.\sin B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right),\,\sin 2A = 2.\sin A.\cos B $ $ $
 $ \sin 3A = 3\sin A - 4{\sin ^3}A,\,\,\sin \left( {\pi - A} \right) = \sin A $

Complete step by step solution:
For convenience we mark given equations as (i), (ii), (iii) and (iv). Therefore we have
 $ 2\tan {10^0} + \tan {50^0} = 2x.......(i) $
 $ \tan {20^0} + \tan {50^0} = 2y.........(ii) $
 $
  2\tan {10^0} + \tan {70^0} = 2w.......(iii) \\
  \tan {20^0} + \tan {70^0} = 2z..........(iv) \;
  $
Subtracting (i) from (ii) we have,
 $
  2y - 2x = \left( {\tan {{20}^0} + \tan {{50}^0}} \right) - \left( {2\tan {{10}^0} + \tan {{50}^0}} \right) \\
   \Rightarrow 2y - 2x = \tan {20^0} + \tan {50^0} - 2\tan {10^0} - \tan {50^0} \\
   \Rightarrow 2y - 2x = \tan {20^0} - 2\tan {10^0} \;
  $
Converting the right hand side of the above equation into a basic trigonometric equation.
i.e. $ $ $ \tan {20^0} = \dfrac{{\sin {{20}^0}}}{{\cos {{20}^0}}},\,\,\tan {10^0} = \dfrac{{\sin {{10}^0}}}{{\cos {{10}^0}}} $ using these in above equation we have,
\[2y - 2x = \dfrac{{\sin {{20}^0}}}{{\cos {{20}^0}}} - 2\left( {\dfrac{{\sin {{10}^0}}}{{\cos {{10}^0}}}} \right)\]
On taking l.c.m. of right hand side
\[ \Rightarrow 2y - 2x = \dfrac{{\left( {\sin {{20}^0}\cos {{10}^0} - 2.\cos {{20}^0}\sin {{10}^0}} \right)}}{{\cos {{20}^0}\cos {{10}^0}}}\]
To make the formula multiply and divide right hand side by $ 2 $ .
 $ 2y - 2x = \dfrac{{2.\sin {{20}^0}\cos {{10}^0} - 4.\cos {{20}^0}\sin {{10}^0}}}{{2\cos {{20}^0}\cos {{10}^0}}} $ or
 $ 2y - 2x = \dfrac{{2.\sin {{20}^0}\cos {{10}^0} - 2\left( {2\cos {{20}^0}\sin {{10}^0}} \right)}}{{2\cos {{20}^0}\cos {{10}^0}}} $
Using identities $ 2\sin A.\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)\,\,and\,\,2\cos A.\sin B = \sin (A + B) - \sin (A - B) $
\[2y - 2x = \dfrac{{\left( {\sin {{30}^0} + \sin {{10}^0}} \right) - 2\left( {\sin {{30}^0} - \sin {{10}^0}} \right)}}{{2\cos {{20}^0}.\cos {{10}^0}}}\]
 $ \Rightarrow 2y - 2x = \dfrac{{\sin {{30}^0} + \sin {{10}^0} - 2.\sin {{30}^0} + 2.\sin {{10}^0}}}{{2\cos {{20}^0}\cos {{10}^0}}} $
 $ \Rightarrow 2y - 2x = \dfrac{{3.\sin {{10}^0} - \sin {{30}^0}}}{{2\cos {{20}^0}\cos {{10}^0}}} $
Writing
  $
  \sin {30^0} = \sin 3\left( {{{10}^0}} \right) \\
  \sin {30^0} = 3\sin \left( {{{10}^0}} \right) - 4{\sin ^3}\left( {{{10}^0}} \right) \\
    \\
  $ or
 $ 3.\sin {10^0} = \sin {30^0} + 4{\sin ^3}{10^0} $
Using the value of $ 3.\sin {10^0} $ in the above equation. We have,
 $ 2y - 2x = \dfrac{{\sin {{30}^0} + 4{{\sin }^3}{{10}^0} - \sin {{30}^0}}}{{2.\cos {{20}^0}.\cos {{10}^0}}} $
 $
   \Rightarrow 2y - 2x = \dfrac{{4{{\sin }^3}{{10}^0}}}{{2\cos {{20}^0}.\cos {{10}^0}}} \\
   \Rightarrow 2y - 2x = \dfrac{{2.{{\sin }^3}{{10}^0}}}{{\cos {{20}^0}.\cos {{10}^0}}} \\
  $
 $ \Rightarrow y - x = \dfrac{{{{\sin }^3}{{10}^0}}}{{\cos {{20}^0}.\cos {{10}^0}}} $
All trigonometric ratios involved in the right hand side of the above equation lie in the first quadrant.
 $ \sin {10^0} > 0,\,\,\cos {20^0} > 0\,\,and\,\,\cos {10^0} > 0 $
Hence, we say that $ y - x > 0 $ or
 $ y > x $ …………………..(A)
Now, subtracting equation (iv) from (iii) we have
 $ 2w - 2z = \left( {2.\tan {{10}^0} + \tan {{70}^0}} \right) - \left( {\tan {{20}^0} + \tan {{70}^0}} \right) $
 $
   \Rightarrow 2w - 2z = 2.\tan {10^0} + \tan {70^0} - \tan {20^0} - \tan {70^0} \\
   \Rightarrow 2w - 2z = 2\tan {10^0} - \tan {20^0} \\
  $
Converting the right hand side of the above equation into a basic trigonometric equation.
i.e. $ $ \[\tan {20^0} = \dfrac{{\sin {{20}^0}}}{{\cos {{20}^0}}},\,\,\tan {10^0} = \dfrac{{\sin {{10}^0}}}{{\cos {{10}^0}}}\] using these in above equation we have
 $ 2w - 2z = 2\left( {\dfrac{{\sin {{10}^0}}}{{\cos {{10}^0}}}} \right) - \left( {\dfrac{{\sin {{20}^0}}}{{\cos {{20}^0}}}} \right) $
On taking l.c.m. of right hand side
 $ 2w - 2z = \dfrac{{2.\cos {{20}^0}\sin {{10}^0} - \sin {{20}^0}.\cos {{10}^0}}}{{\cos {{10}^0}.\cos {{20}^0}}} $
To make the formula multiply and divide right hand side by $ 2 $ .
 $ 2w - 2z = \dfrac{{4.\cos {{20}^0}\sin {{10}^0} - 2.\sin {{20}^0}.\cos {{10}^0}}}{{2.cos{{10}^0}.\cos {{20}^0}}} $ or
 $ 2w - 2z = \dfrac{{2\left( {2.\cos {{20}^0}\sin {{10}^0}} \right) - \left( {2.\sin {{20}^0}.\cos {{10}^0}} \right)}}{{2.cos{{10}^0}.\cos {{20}^0}}} $
Using identities $ 2\sin A.\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)\,\,and\,\,2\cos A.\sin B = \sin (A + B) - \sin (A - B) $
 $ 2w - 2z = \dfrac{{2\left( {\sin {{30}^0} - \sin {{10}^0}} \right) - \left( {\sin {{30}^0} + \sin {{10}^0}} \right)}}{{2.\cos {{10}^0}\cos {{20}^0}}} $
 $ \Rightarrow 2w - 2z = \dfrac{{2.\sin {{30}^0} - 2.\sin {{10}^0} - \sin {{30}^0} - \sin {{10}^0}}}{{2.\cos {{10}^0}\cos {{20}^0}}} $
 $ \Rightarrow 2w - 2z = \dfrac{{\sin {{30}^0} - 3\sin {{10}^0}}}{{2.\cos {{10}^0}\cos {{20}^0}}} $
Writing
  $
  \sin {30^0} = \sin 3\left( {{{10}^0}} \right) \\
  \sin {30^0} = 3\sin \left( {{{10}^0}} \right) - 4{\sin ^3}\left( {{{10}^0}} \right) \\
  $
Using, value of $ \sin {30^0} $ in the above equation. We have,

 $ 2w - 2z = \dfrac{{3.sin{{10}^0} - 4.{{\sin }^3}{{10}^0} - 3.\sin {{10}^0}}}{{2.\cos {{10}^0}.\cos {{20}^0}}} $
 $
   \Rightarrow 2w - 2z = \dfrac{{ - 4.{{\sin }^3}{{10}^0}}}{{2.\cos {{10}^0}.\cos {{20}^0}}} \;
   \Rightarrow 2w - 2z = \dfrac{{ - 2.{{\sin }^3}{{10}^0}}}{{\cos {{10}^0}.\cos {{20}^0}}} \\
   \Rightarrow z - w = \dfrac{{{{\sin }^3}{{10}^0}}}{{\cos {{10}^0}.\cos {{20}^0}}} \\
  $
All trigonometric ratios involved in the right hand side of the above equation lies in the first quadrant.
 $ \sin {10^0} > 0,\,\,\cos {20^0} > 0\,\,and\,\,\cos {10^0} > 0 $
Hence, we say that $ y - x > 0 $ or
 $ z > w $ …………………..(B)
Now, subtracting (ii) from (iii) we have
 $ 2w - 2y = \left( {2\tan {{10}^0} + \tan {{70}^0}} \right) - \left( {\tan {{20}^0} + \tan {{50}^0}} \right) $
 $
   \Rightarrow 2w - 2y = 2x \\
   \Rightarrow w - y = x \\
  $
Or we can write
 $ w = x + y $ ……(C)
 $ \Rightarrow w > y $ …….(D)
Then from equations (A), (B) and (D) we have
 $ z > w > y > x $
Therefore, option (A) is the correct option.
Now, we will discuss others options also,
From equation (C) derived above we have
 $ w = x + y $
Hence, option (B) is also correct.
To discuss third option we consider
 $ 2y - z = \tan {20^0} + \tan {50^0} - \dfrac{1}{2}\tan {20^0} - \dfrac{1}{2}\tan {70^0} $
 $
   \Rightarrow 2y - z = \dfrac{1}{2}\tan {20^0} + \tan {50^0} - \dfrac{1}{2}\tan {70^0} \\
   \Rightarrow 2y - z = \dfrac{1}{2}\left( {\tan {{20}^0} + 2\tan {{50}^0} - \tan {{70}^0}} \right) \;
  $
 $ \Rightarrow 2y - z = \dfrac{1}{2}\left( {\dfrac{{\sin {{70}^0}}}{{\cos {{20}^0}\cos {{50}^0}}} - \dfrac{{\sin {{20}^0}}}{{\cos {{50}^0}\cos {{70}^0}}}} \right) $
 $ \Rightarrow 2y - z = \dfrac{1}{2}\dfrac{{\left( {\sin {{70}^0}\cos {{70}^0} - \sin {{20}^0}.\cos {{20}^0}} \right)}}{{\cos {{20}^0}.\cos {{50}^0}.\cos {{70}^0}}} $
Multiply and divide by 2 on the right hand side of the above equation.
 $ 2y - z = \dfrac{1}{4}.\dfrac{{2.\sin {{70}^0}\sin {{70}^0} - 2\sin {{20}^0}\cos {{20}^0}}}{{\cos {{20}^0}.\cos {{50}^0}\cos {{70}^0}}} $
 $ \Rightarrow 2y - z = \dfrac{1}{4}.\dfrac{{\sin {{140}^0} - \sin {{40}^0}}}{{\cos {{20}^0}.\cos {{50}^0}.\cos {{70}^0}}} $
 $ $ $ \Rightarrow 2y - z = \dfrac{1}{4}.\dfrac{{\sin {{40}^0} - \sin {{40}^0}}}{{\cos {{20}^0}.\cos {{50}^0}\cos {{70}^0}}} $
 $ $ $
   \Rightarrow 2y - z = 0 \\
   \Rightarrow 2y = z \;
  $
Hence, from above we see that option (C) is also correct.
Now, for fourth option we see that
 $ w = x + y $ and $ w > y $
Hence, $ z + x = w + y $ is not possible.
Therefore, option (D) is incorrect.
Hence, from above we see that option (A), (B) and (C) all three are correct options.
So, the correct answer is “Option A ,B AND C”.

Note: When a number of equations are given in a problem first we group them having common terms and then using appropriate formulas to solve them or to drive a relation between given variables. This can be only done by choosing the right grouping of equations.