If A = 10°, what is the value of: $$\frac{{12\sin 3A + 5\cos \left( {5A - {5^ \circ }} \right)}}{{9\sin \frac{{9A}}{2} - 4\cos \left( {5A + {{10}^ \circ }} \right)}}?$$
A. $$\frac{{6\sqrt 2 + 5}}{{9 - 2\sqrt 2 }}$$
B. $$\frac{{6\sqrt 2 - 5}}{{9 - 2\sqrt 2 }}$$
C. $$\frac{{9 - 2\sqrt 2 }}{{6\sqrt 2 + 5}}$$
D. $$\frac{{6\sqrt 2 + 5}}{{9 + 2\sqrt 2 }}$$
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & A = {10^ \circ } \cr & \frac{{12\sin 3A + 5\cos \left( {5A - {5^ \circ }} \right)}}{{9\sin \frac{{9A}}{2} - 4\cos \left( {5A + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin 3 \times {{10}^ \circ } + 5\cos \left( {5 \times {{10}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{9 \times {{10}^ \circ }}}{2} - 4\cos \left( {5 \times {{10}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos \left( {{{50}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{{{90}^ \circ }}}{2} - 4\cos \left( {{{50}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos {{45}^ \circ }}}{{9\sin {{45}^ \circ } - 4\cos {{60}^ \circ }}} \cr & = \frac{{12 \times \frac{1}{2} + 5 \times \frac{1}{{\sqrt 2 }}}}{{9 \times \frac{1}{{\sqrt 2 }} - 4 \times \frac{1}{2}}} \cr & = \frac{{6 + \frac{5}{{\sqrt 2 }}}}{{\frac{9}{{\sqrt 2 }} - 2}} \cr & = \frac{{6\sqrt 2 + 5}}{{9 - 2\sqrt 2 }} \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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