Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^4}\theta = 1 - {\cos ^2}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^4}\theta = {\sin ^2}\theta \cr & \Rightarrow {\cos ^2}\theta .{\cos ^2}\theta = {\sin ^2}\theta \cr & \Rightarrow {\cos ^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{ta}}{{\text{n}}^4}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr} $$

$$\frac{{3\sin \theta + 2{\text{cos}}\theta }}{{3\sin \theta - 2{\text{cos}}\theta }}$$
Divide numerator & denominator by cosθ
$$\eqalign{ & = \frac{{\frac{{3\sin \theta }}{{\cos \theta }} + \frac{{2\cos \theta }}{{\cos \theta }}}}{{\frac{{3\sin \theta }}{{\cos \theta }} - \frac{{2\cos \theta }}{{\cos \theta }}}}\left[ {\frac{{\sin \theta }}{{\cos \theta }} = \tan \theta } \right] \cr & = \frac{{3\tan \theta + 2}}{{3\tan \theta - 2}} \cr & {\text{Put value of tan}}\theta \cr & = \frac{{3 \times \frac{4}{3} + 2}}{{3 \times \frac{4}{3} - 2}} \cr & = \frac{6}{2} \cr & = 3 \cr} $$

(secA - cosA)² + (cosecA - sinA)² - (cotA - tanA)²
= (sec²A + cos²A - 2secA.cosA) + (coses²A + sin²A - 2cosecA.sinA) - (cot²A + tan²A - 2cotA.tanA)
= sec²A - tan²A + cos²A + sin²A + coses²A - cot²A - 2
= 3 - 2
= 1

Alternate shortcut method:
(secA - cosA)² + (cosecA - sinA)² - (cotA - tanA)²
Put θ = 45°
= (sec45° - cos45°)² + (cosec45° - sin45°)² - (cot45° - tan45°)²
$$\eqalign{ & = {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} + {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} - {\left( {1 - 1} \right)^2} \cr & = \frac{1}{2} + \frac{1}{2} - 0 \cr & = 1 \cr} $$

(x + 5)° + (2x - 3)° + (3x + 4)° = 180°
(Sum of all angles in triangle is 180°)
⇒ 6x + 6° = 180°
⇒ (x + 1) = 30°
⇒ x = 29°

$$\eqalign{ & \frac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}} \cr & = \frac{{cot{{33}^ \circ } + \tan {{53}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}} \cr & = \frac{{\frac{1}{{\tan {{33}^ \circ }}} + \tan {{53}^ \circ }}}{{\tan {{33}^ \circ } + \frac{1}{{\tan {{53}^ \circ }}}}} \cr & = \frac{{1 + \tan {{53}^ \circ }.\tan {{33}^ \circ }}}{{\tan {{33}^ \circ }.\tan {{53}^ \circ } + 1}} \times \frac{{\tan {{53}^ \circ }}}{{\tan {{33}^ \circ }}} \cr & = \tan {53^ \circ }.cot{33^ \circ } \cr & = \cot {37^ \circ }.\tan {57^ \circ } \cr} $$

$$\eqalign{ & \frac{{\cot {{30}^ \circ } - \cot {{75}^ \circ }}}{{\tan {{15}^ \circ } - \tan {{60}^ \circ }}} \cr & = \frac{{\tan {{60}^ \circ } - \tan {{15}^ \circ }}}{{\tan {{15}^ \circ } - \tan {{60}^ \circ }}} \cr & = \frac{{ - \left( {\tan {{15}^ \circ } - \tan {{60}^ \circ }} \right)}}{{\tan {{15}^ \circ } - \tan {{60}^ \circ }}} \cr & = - 1 \cr} $$

$$\eqalign{ & \left( {{\text{se}}{{\text{c}}^4}\theta - {\text{ta}}{{\text{n}}^4}\theta } \right) \cr & \Rightarrow \left( {{\text{se}}{{\text{c}}^2}\theta - {\text{ta}}{{\text{n}}^2}\theta } \right)\left( {{\text{se}}{{\text{c}}^2}\theta + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow 1 \times \left( {{\text{se}}{{\text{c}}^2}\theta + {\text{ta}}{{\text{n}}^2}\theta } \right)[1 + {\text{ta}}{{\text{n}}^2}\theta = {\text{se}}{{\text{c}}^2}\theta ] \cr & \Rightarrow 1 \times \frac{7}{{12}} \cr & \Rightarrow \frac{7}{{12}} \cr} $$

$$\eqalign{ & \frac{5}{{{\text{se}}{{\text{c}}^2}\theta }}{\text{ + }}\frac{2}{{1 + {\text{co}}{{\text{t}}^2}\theta }}{\text{ + 3}}{\sin ^2}\theta \cr & = 5{\cos ^2}\theta + \frac{2}{{{\text{cose}}{{\text{c}}^2}\theta }} + 3{\sin ^2}\theta \cr & = 5{\text{co}}{{\text{s}}^2}\theta + 2{\sin ^2}\theta + 3{\sin ^2}\theta \cr & = 5\left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right) \cr & \left( {\because {{\sin }^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1} \right) \cr & = 5 \times 1 \cr & = 5 \cr} $$

$$\eqalign{ & \left( {\frac{1}{{\cos \theta }} + \frac{1}{{\cot \theta }}} \right){\text{ }}\left( {\frac{1}{{\cos \theta }} - \frac{1}{{\cot \theta }}} \right) \cr & = \left( {\sec \theta + \tan \theta } \right)\left( {\sec \theta - \tan \theta } \right) \cr & = {\sec ^2}\theta - {\tan ^2}\theta \left[ {1 + {{\tan }^2}\theta = {{\sec }^2}\theta } \right] \cr & = 1 \cr} $$

cos1° cos2° cos3° ............. cos177° cos178° cos179°
= cos90°
= 0[0 will make whole series 0]
= 0