Question

Let \[\sin \left( {\alpha - \beta } \right) = \dfrac{5}{{13}}\]and \[\cos \left( {\alpha + \beta } \right) = \dfrac{3}{5}\] where \[\alpha ,\beta \in \left( {0,\dfrac{\pi }{4}} \right)\] then \[\tan 2\alpha \]
A. \[\dfrac{{63}}{{16}}\]
B. \[\dfrac{{61}}{{16}}\]
C. \[\dfrac{{65}}{{16}}\]
D. \[\dfrac{{32}}{9}\]

Answer 1

Hint: In the given question, we need to find the value of \[\tan 2\alpha \]. For that, we use the basic trigonometric formulas like \[{\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}}\] and identities like \[\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}\] to get the desired result.

Formula used:
We have been using the following formulas:
1. \[{\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}}\]
2. \[{\sec ^2}x = 1 + {\tan ^2}x\]
3. \[{\sin ^2}x + {\cos ^2}x = 1\]
4. \[\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}\]

Complete step-by-step solution:
Given that
\[\cos \left( {\alpha + \beta } \right) = \dfrac{3}{5}...\left( 1 \right)\]
\[\sin \left( {\alpha - \beta } \right) = \dfrac{5}{{13}}...\left( 2 \right)\]
Now take squares on both sides in equation (1):
\[
  {\cos ^2}\left( {\alpha + \beta } \right) = {\left( {\dfrac{3}{5}} \right)^2} \\
   = \dfrac{9}{{25}}
 \]
Now take the inverse of both sides of the above equation:
\[
  \dfrac{1}{{{{\cos }^2}\left( {\alpha + \beta } \right)}} = \dfrac{1}{{\dfrac{9}{{25}}}} \\
   = \dfrac{{25}}{9}...\left( 3 \right)
 \]
We know that \[{\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}}\]
Now apply this formula to equation (3):
\[{\sec ^2}\left( {\alpha + \beta } \right) = \dfrac{{25}}{9}...\left( 4 \right)\]
We know that \[{\sec ^2}x = 1 + {\tan ^2}x\]
Now apply this identity to equation (4):
\[
  1 + {\tan ^2}\left( {\alpha + \beta } \right) = \dfrac{{25}}{9} \\
  {\tan ^2}\left( {\alpha + \beta } \right) = \dfrac{{25}}{9} - 1 \\
  {\tan ^2}\left( {\alpha + \beta } \right) = \dfrac{{25 - 9}}{9} \\
   = \dfrac{{16}}{9}
 \]
Further solving,
\[{\tan ^2}\left( {\alpha + \beta } \right) = {\left( {\dfrac{4}{3}} \right)^2}\]
By canceling squares on both sides:
\[\tan \left( {\alpha + \beta } \right) = \pm \dfrac{4}{3}\]
Given that \[\alpha ,\beta \in \left( {0,\dfrac{\pi }{4}} \right)\] that means
\[
  0 \leqslant \alpha \leqslant \dfrac{\pi }{4}...\left( 5 \right) \\
  0 \leqslant \beta \leqslant \dfrac{\pi }{4}...\left( 6 \right)
 \]
Now add equations (5) and (6):
\[
  0 + 0 \leqslant \alpha + \beta \leqslant \dfrac{\pi }{4} + \dfrac{\pi }{4} \\
  0 \leqslant \alpha + \beta \leqslant \dfrac{{\pi + \pi }}{4} \\
  0 \leqslant \alpha + \beta \leqslant \dfrac{{2\pi }}{4} \\
  0 \leqslant \alpha + \beta \leqslant \dfrac{\pi }{2}
 \]
Thus, tan is positive in the first quadrant.
Therefore, \[\tan \left( {\alpha + \beta } \right) = \dfrac{4}{3}\]
Consider \[\sin \left( {\alpha - \beta } \right) = \dfrac{5}{{13}}...\left( 2 \right)\]
Take squares on both sides of equation (2):
\[
  {\sin ^2}\left( {\alpha - \beta } \right) = {\left( {\dfrac{5}{{13}}} \right)^2} \\
   = \dfrac{{25}}{{169}}
 \]
We know that \[{\sin ^2}x + {\cos ^2}x = 1\]
By applying this formula to the above equation:
\[\dfrac{{25}}{{169}} + {\cos ^2}\left( {\alpha - \beta } \right) = 1\]
By simplifying, we get
\[
  {\cos ^2}\left( {\alpha - \beta } \right) = 1 - \dfrac{{25}}{{169}} \\
   = \dfrac{{169 - 25}}{{169}} \\
   = \dfrac{{144}}{{169}}
 \]
By taking the inverse on both sides of the above equation:
\[
  \dfrac{1}{{{{\cos }^2}\left( {\alpha - \beta } \right)}} = \dfrac{1}{{\dfrac{{144}}{{169}}}} \\
   = \dfrac{{169}}{{144}}...\left( 7 \right)
 \]
We know that \[{\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}}\]
By applying this formula in equation (7):
\[{\sec ^2}\left( {\alpha - \beta } \right) = \dfrac{{169}}{{144}}\]
We know that \[{\sec ^2}x = 1 + {\tan ^2}x\]
By applying this identity to our equation:
\[
  1 + {\tan ^2}\left( {\alpha - \beta } \right) = \dfrac{{169}}{{144}} \\
  {\tan ^2}\left( {\alpha - \beta } \right) = \dfrac{{169}}{{144}} - 1 \\
   = \dfrac{{169 - 144}}{{144}} \\
   = \dfrac{{25}}{{144}}
 \]
Further simplifying,
\[\tan \left( {\alpha - \beta } \right) = \pm \dfrac{5}{{12}}\]
Now the angle \[\alpha - \beta \] lies in the first quadrant.
Thus, tan is positive.
\[\tan \left( {\alpha - \beta } \right) = \dfrac{5}{{12}}\]
Now we know that \[\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}\]
Therefore, \[\tan \left( {\alpha + \beta + \left( {\alpha - \beta } \right)} \right) = \dfrac{{\tan \left( {\alpha + \beta } \right) + \tan \left( {\alpha - \beta } \right)}}{{1 - \tan \left( {\alpha + \beta } \right)\tan \left( {\alpha - \beta } \right)}}\]
By substituting all the values in it:
\[
  \tan \left( {\alpha + \beta + \alpha - \beta } \right) = \dfrac{{\dfrac{4}{3} + \dfrac{5}{{12}}}}{{1 - \dfrac{4}{3} \times \dfrac{5}{{12}}}} \\
  \tan \left( {2\alpha } \right) = \dfrac{{\dfrac{{16 + 5}}{{12}}}}{{1 - \dfrac{{20}}{{36}}}} \\
   = \dfrac{{\dfrac{{21}}{{12}}}}{{\dfrac{{36 - 20}}{{36}}}} \\
   = \dfrac{{\dfrac{{21}}{{12}}}}{{\dfrac{{16}}{{36}}}}
 \]
Further simplifying,
\[
  \tan \left( {2\alpha } \right) = \dfrac{{21}}{{12}} \times \dfrac{{36}}{{16}} \\
   = \dfrac{{21 \times 3}}{{16}} \\
   = \dfrac{{63}}{{16}}
\]
Hence, option (A) is correct

Note: Students must know the properties of trigonometric formulas to solve this type of question, and they should be careful while taking the inverse of the function, as it has a chance of making a mistake.