Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & \frac{x}{8} = \sin \theta + \cos \theta ........\left( {\text{i}} \right) \cr & \frac{y}{9} = \sin \theta - \cos \theta ........\left( {{\text{ii}}} \right) \cr & {\text{Square and add equation}}\left( {\text{i}} \right){\text{and}}\left( {{\text{ii}}} \right) \cr & \frac{{{x^2}}}{{{8^2}}} + \frac{{{y^2}}}{{{9^2}}} = {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta + {\sin ^2}\theta + {\cos ^2}\theta - 2\sin \theta \cos \theta \cr & \frac{{{x^2}}}{{{8^2}}} + \frac{{{y^2}}}{{{9^2}}} = 2 \cr} $$

$$\eqalign{ & \frac{{\left( {\cos {9^ \circ } + \sin {{81}^ \circ }} \right)\left( {\sec {9^ \circ } + {\text{cosec}}\,{\text{8}}{{\text{1}}^ \circ }} \right)}}{{{\text{cose}}{{\text{c}}^2}71 + {{\cos }^2}{{15}^ \circ } - {{\tan }^2}{{19}^ \circ } + {{\cos }^2}{{75}^ \circ }}} \cr & = \frac{{\left( {\cos {9^ \circ } + \cos {9^ \circ }} \right)\left( {\sec {9^ \circ } + \sec {9^ \circ }} \right)}}{{{\text{cose}}{{\text{c}}^2}71 + {{\cos }^2}{{15}^ \circ } - {{\cot }^2}{{71}^ \circ } + {{\sin }^2}{{15}^ \circ }}} \cr & = \frac{{\left( {2\cos {9^ \circ }} \right)\left( {2\sec {9^ \circ }} \right)}}{{1 + 1}} \cr & = 2 \cr} $$

$$\eqalign{ & \tan \theta + \sec \theta = \frac{{x - 2}}{{x + 2}}........\left( {\text{i}} \right) \cr & \frac{{\left( {\sec \theta + \tan \theta } \right)}}{1} \times \frac{{\left( {\sec \theta - \tan \theta } \right)}}{{\left( {\sec \theta - \tan \theta } \right)}} = \frac{{x - 2}}{{x + 2}} \cr & \frac{1}{{\sec \theta - \tan \theta }} = \frac{{x - 2}}{{x + 2}} \cr & \left( {\because \,{{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right) \cr & \sec \theta - \tan \theta = \frac{{x + 2}}{{x - 2}}........\left( {{\text{ii}}} \right) \cr & {\text{By equation }}\left( {{\text{ii}}} \right) + \left( {\text{i}} \right){\text{ we get}} \cr & 2\sec \theta = \frac{{x + 2}}{{x - 2}} + \frac{{x - 2}}{{x + 2}} \cr & 2\sec \theta = \frac{{{{\left( {x + 2} \right)}^2} + {{\left( {x - 2} \right)}^2}}}{{\left( {{x^2} - {2^2}} \right)}} \cr & 2\sec \theta = \frac{{\left[ {{x^2} + 4 + 2x + {x^2} + 4 - 2x} \right]}}{{\left( {{x^2} - 4} \right)}} \cr & 2\sec \theta = \frac{{2\left( {{x^2} + 4} \right)}}{{\left( {{x^2} - 4} \right)}} \cr & \sec \theta = \frac{{{x^2} + 4}}{{{x^2} - 4}} \cr & \frac{1}{{\cos \theta }} = \frac{{{x^2} + 4}}{{{x^2} - 4}} \cr & \cos \theta = \frac{{{x^2} - 4}}{{{x^2} + 4}} \cr} $$

cosA + cosB + cosC = 3
⇒ putting A, B, C = 0
1 + 1 + 1 = 3
3 = 3 Satisfied
So, sinA + sinB + sinC
= sin0° + sin0° + sin0°
= 0 + 0 + 0
= 0

$$\eqalign{ & \frac{{\cos x - \sqrt 3 \sin x}}{2} \cr & \Rightarrow \frac{1}{2}\cos x - \frac{{\sqrt 3 }}{2}\sin x \cr & \Rightarrow \cos {60^ \circ }.\cos x - \sin {60^ \circ }.\sin x \cr & \Rightarrow \cos \left( {{{60}^ \circ } + x} \right) \cr & \Rightarrow \cos \left( {\frac{\pi }{3} + x} \right) \cr} $$

$$\eqalign{ & \left( {1 + {{\cot }^2}\theta } \right)\left( {1 + {{\tan }^2}\theta } \right) \times \left( {\sin \theta - {\text{cosec}}\,\theta } \right)\left( {\cos \theta - \sec \theta } \right) \cr & = \left( {{\text{cose}}{{\text{c}}^2}\theta } \right)\left( {{{\sec }^2}\theta } \right) \times \left( {\frac{{{{\sin }^2}\theta - 1}}{{\sin \theta }}} \right)\left( {\frac{{{{\cos }^2}\theta - 1}}{{\cos \theta }}} \right) \cr & = \left( {{\text{cose}}{{\text{c}}^2}\theta } \right)\left( {{{\sec }^2}\theta } \right) \times \left( {\frac{{ - {{\cos }^2}\theta }}{{\sin \theta }}} \right)\left( {\frac{{ - {{\sin }^2}\theta }}{{\cos \theta }}} \right) \cr & = \left( {{\text{cose}}{{\text{c}}^2}\theta } \right)\left( {{{\sec }^2}\theta } \right) \times \sin \theta \cos \theta \cr & = \sec \theta \,{\text{cosec}}\,\theta \cr} $$

$$\eqalign{ & \frac{{\left( {1 - 2{{\sin }^2}\theta {{\cos }^2}\theta } \right)\left( {\cot \theta + 1} \right)\cos \theta }}{{\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right)\left( {1 + \tan \theta } \right){\text{cosec}}\,\theta }} - 1 \cr & {\text{Put }}\theta = {45^ \circ } \cr & = \frac{{\left( {1 - 2 \times {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right)\left( {1 + 1} \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^4} + {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^4}} \right)\left( {1 + 1} \right)\sqrt 2 }} - 1 \cr & = \frac{{\left( {1 - 2 \times \frac{1}{4}} \right)\left( 2 \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {\frac{1}{4} + \frac{1}{4}} \right)\left( {1 + 1} \right)\sqrt 2 }} - 1 \cr & = \frac{{\left( {\frac{1}{2}} \right)\left( 2 \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {\frac{1}{2}} \right)\left( 2 \right)\sqrt 2 }} - 1 \cr & = \frac{{\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\sqrt 2 }} - 1 \cr & = \frac{1}{2} - 1 \cr & = - \frac{1}{2} \cr & {\text{Option C is answer}} \cr & \Rightarrow - {\sin ^2}\theta \cr & = - {\sin ^2}{45^ \circ } \cr & = - {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & = - \frac{1}{2} \cr} $$

$$\eqalign{ & \frac{{\cos \left( {a - b} \right) - \cos \left( {a + b} \right)}}{{\cos \left( {a - b} \right) + \cos \left( {a + b} \right)}} \cr & = \frac{{\cos \left( {{{45}^ \circ } - {{15}^ \circ }} \right) - \cos \left( {{{45}^ \circ } + {{15}^ \circ }} \right)}}{{\cos \left( {{{45}^ \circ } - {{15}^ \circ }} \right) + \cos \left( {{{45}^ \circ } + {{15}^ \circ }} \right)}} \cr & = \frac{{\cos {{30}^ \circ } - \cos {{60}^ \circ }}}{{\cos {{30}^ \circ } + \cos {{60}^ \circ }}} \cr & = \frac{{\frac{{\sqrt 3 }}{2} - \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2} + \frac{1}{2}}} \cr & = \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \times \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}} \cr & = \frac{{3 + 1 - 2\sqrt 3 }}{2} \cr & = \frac{{4 - 2\sqrt 3 }}{2} \cr & = 2 - \sqrt 3 \cr} $$

$$\eqalign{ & \cos \left( { - \frac{{17\pi }}{3}} \right) \cr & = \cos \left( { - \frac{{17 \times 180}}{3}} \right) \cr & = \cos \left( { - 1020} \right) \cr & = \cos \left[ {3 \times 360 - 60} \right] \cr & = \cos 60 \cr & = \frac{1}{2} \cr} $$

\[\begin{array}{l} \frac{{\left[ {1 - \tan \left( {90 - \theta } \right) + \sec \left( {90 - \theta } \right)} \right]}}{{\left[ {\tan \left( {90 - \theta } \right) - \sec \left( {90 - \theta } \right) + 1} \right]}}\\ \Rightarrow \frac{{\left[ {1 - \cot \theta + {\rm{cosec}}\,\theta } \right]}}{{\left[ {\cot \theta + {\rm{cosec}}\,\theta + 1} \right]}}\\ \Rightarrow \frac{{\left[ {1 - \frac{{\cos \theta }}{{\sin \theta }} + \frac{1}{{\sin \theta }}} \right]}}{{\left[ {\frac{{\cos \theta }}{{\sin \theta }} + \frac{1}{{\sin \theta }} + 1} \right]}}\\ \Rightarrow \frac{{\left[ {\sin \theta - \cos \theta + 1} \right]}}{{\left[ {\sin \theta + \cos \theta + 1} \right]}}\\ \Rightarrow \frac{{\left( {\sin \theta + 1} \right) - \cos \theta }}{{\left( {\sin \theta + 1} \right) + \cos \theta }}\\ \left[ \begin{array}{l} \therefore \sin \theta = 2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}\\ \cos \theta = 1 - 2{\sin ^2}\frac{\theta }{2}\\ \cos \frac{\theta }{2} = 2{\cos ^2}\frac{\theta }{2} - 1 \end{array} \right]\\ \Rightarrow \frac{{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2} + 1 - 1 + 2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}.\cos \frac{\theta }{2} + 1 + 2{{\cos }^2}\frac{\theta }{2} - 1}}\\ \Rightarrow \frac{{2\sin \frac{\theta }{2}\left( {\sin \frac{\theta }{2} + \cos \frac{\theta }{2}} \right)}}{{2\cos \frac{\theta }{2}\left( {\sin \frac{\theta }{2} + \cos \frac{\theta }{2}} \right)}}\\ \Rightarrow \tan \frac{\theta }{2} \end{array}\]