If a = 45° and b = 15°, what is the value of $$\frac{{\cos \left( {a - b} \right) - \cos \left( {a + b} \right)}}{{\cos \left( {a - b} \right) + \cos \left( {a + b} \right)}}?$$
A. 2 - 2√2
B. 3 - √6
C. 3 - √2
D. 2 - √3
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & \frac{{\cos \left( {a - b} \right) - \cos \left( {a + b} \right)}}{{\cos \left( {a - b} \right) + \cos \left( {a + b} \right)}} \cr & = \frac{{\cos \left( {{{45}^ \circ } - {{15}^ \circ }} \right) - \cos \left( {{{45}^ \circ } + {{15}^ \circ }} \right)}}{{\cos \left( {{{45}^ \circ } - {{15}^ \circ }} \right) + \cos \left( {{{45}^ \circ } + {{15}^ \circ }} \right)}} \cr & = \frac{{\cos {{30}^ \circ } - \cos {{60}^ \circ }}}{{\cos {{30}^ \circ } + \cos {{60}^ \circ }}} \cr & = \frac{{\frac{{\sqrt 3 }}{2} - \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2} + \frac{1}{2}}} \cr & = \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \times \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}} \cr & = \frac{{3 + 1 - 2\sqrt 3 }}{2} \cr & = \frac{{4 - 2\sqrt 3 }}{2} \cr & = 2 - \sqrt 3 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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