Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & 4{\sin ^2}{30^ \circ } + 3{\cot ^2}{60^ \circ } \cr & = 4 \times {\left( {\frac{1}{2}} \right)^2} + 3{\left( {\frac{1}{{\sqrt 3 }}} \right)^2} \cr & = 4 \times \frac{1}{4} + 3 \times \frac{1}{3} \cr & = 1 + 1 \cr & = 2 \cr} $$

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

$$\eqalign{ & 5{\sin ^2}{60^ \circ } + 7{\sin ^2}{45^ \circ } + 8{\cos ^2}{45^ \circ } \cr & = 5{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + 7{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 8{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \cr & = \frac{{15}}{4} + \frac{7}{2} + \frac{8}{2} \cr & = \frac{{15 + 14 + 16}}{4} \cr & = \frac{{45}}{4} \cr} $$

$$\eqalign{ & \frac{{\cos {{40}^ \circ } - \cos {{140}^ \circ }}}{{\sin {{80}^ \circ } + \sin {{20}^ \circ }}} \cr & = \frac{{ - 2\sin \left( {\frac{{{{40}^ \circ } + {{140}^ \circ }}}{2}} \right)\sin \left( {\frac{{{{40}^ \circ } - {{140}^ \circ }}}{2}} \right)}}{{2\sin \left( {\frac{{{{80}^ \circ } + {{20}^ \circ }}}{2}} \right)\cos \left( {\frac{{{{80}^ \circ } - {{20}^ \circ }}}{2}} \right)}} \cr & = \frac{{ - 2\sin \left( {\frac{{{{180}^ \circ }}}{2}} \right)\sin \left( { - \frac{{{{100}^ \circ }}}{2}} \right)}}{{2\sin \left( {\frac{{{{100}^ \circ }}}{2}} \right)\cos \left( {\frac{{{{60}^ \circ }}}{2}} \right)}} \cr & = \frac{{ - 2\sin {{90}^ \circ } \times \sin \left( { - {{50}^ \circ }} \right)}}{{2\sin {{50}^ \circ } \times \cos {{30}^ \circ }}} \cr & = \frac{{ - 2\sin {{90}^ \circ } \times \left( { - \sin {{50}^ \circ }} \right)}}{{2\sin {{50}^ \circ } \times \cos {{30}^ \circ }}} \cr & = \frac{{2\sin {{90}^ \circ } \times \sin {{50}^ \circ }}}{{2\sin {{50}^ \circ } \times \cos {{30}^ \circ }}} \cr & = \frac{{\sin {{90}^ \circ }}}{{\cos {{30}^ \circ }}} \cr & = \frac{1}{{\frac{{\sqrt 3 }}{2}}} \cr & = \frac{2}{{\sqrt 3 }} \cr} $$

(sinθ + cosecθ)² + (cosθ + secθ)² - 1
= sin²θ + cosec²θ + 2 + cos²θ + sec²θ + 2 - 1
= (sin²θ + cos²θ) + 1 + cot²θ + 2 + 1 + tan²θ + 2 - 1
= 6 + tan²θ + cot²θ

sec4θ = cosec(θ + 20°)
sec4θ = sec[90° - (θ + 20°)]
sec4θ = sec(70° - θ)
4θ = 70° - θ
5θ = 70°
θ = 14°

cos2Acos2B + sin²(A - B) - sin²(A + B)
= cos2Acos2B + [{sin(A - B) + sin(A + B)}{sin(A - B) - sin(A + B)}]
= cos2Acos2B + [(sinAcosB - cosAsinB + sinAcosB + cosAsinB)(sinAcosB - cosAsinB - sinAcosB - cosAsinB)]
= cos2Acos2B + [(2sinAcosB) × (-2cosAsinB)]
= cos2Acos2B - (2sinAcosA) × (2sinBcosB)
= cos2Acos2B - sin2Asin2B
= cos(2A + 2B)

Alternate:
cos2Acos2B + sin²(A - B) - sin²(A + B)
= cos2Acos2B + sin2Asin2B
= cos(2A + 2B)

$$\eqalign{ & \frac{{\left( {\cos {9^ \circ } + \sin {{81}^ \circ }} \right)\left( {\sec {9^ \circ } + {\text{cosec}}\,{{81}^ \circ }} \right)}}{{\sin {{56}^ \circ }\sec {{34}^ \circ } + \cos {{25}^ \circ }{\text{cosec}}\,{{65}^ \circ }}} \cr & = \frac{{\left( {\cos {9^ \circ } + \cos {9^ \circ }} \right)\left( {\sec {9^ \circ } + \sec {9^ \circ }} \right)}}{{\cos {{34}^ \circ }\sec {{34}^ \circ } + \cos {{25}^ \circ }\sec {{25}^ \circ }}} \cr & = \frac{{2\cos {9^ \circ } \times 2\sec {9^ \circ }}}{{1 + 1}} \cr & = \frac{{4 \times \cos {9^ \circ }\sec {9^ \circ }}}{2} \cr & = 2 \cr} $$

$$\eqalign{ & 7{\sin ^2}\theta + 4{\cos ^2}\theta = 5 \cr & 3{\sin ^2}\theta + 4{\sin ^2}\theta + 4{\cos ^2}\theta = 5 \cr & 3{\sin ^2}\theta + 4 = 5 \cr & 3{\sin ^2}\theta = 5 - 4 \cr & 3{\sin ^2}\theta = 1 \cr & {\sin ^2}\theta = \frac{1}{3} \cr & \sin \theta = \frac{1}{{\sqrt 3 }} = \frac{P}{H} \cr & B = \sqrt {{{\left( {\sqrt 3 } \right)}^2} - {1^2}} = \sqrt 2 \cr & \Rightarrow \frac{{\sqrt 3 \sec \theta + \tan \theta }}{{\sqrt 2 \cot \theta - \sqrt 3 \cos \theta }} \cr & = \frac{{\sqrt 3 \times \frac{{\sqrt 3 }}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}}}{{\sqrt 2 \times \frac{{\sqrt 2 }}{1} - \sqrt 3 \times \frac{{\sqrt 2 }}{{\sqrt 3 }}}} \cr & = \frac{{\frac{4}{{\sqrt 2 }}}}{{2 - \sqrt 2 }} \cr & = \frac{{2\sqrt 2 }}{{2 - \sqrt 2 }} \cr & = \frac{{2\sqrt 2 }}{{2 - \sqrt 2 }} \times \frac{{2 + \sqrt 2 }}{{2 + \sqrt 2 }} \cr & = \frac{{2\sqrt 2 \left( {2 + \sqrt 2 } \right)}}{{4 - 2}} \cr & = 2\left( {\sqrt 2 + 1} \right) \cr} $$