Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

12cot²θ - 31cosecθ + 32 = 0
12cosec²θ - 12 - 31cosecθ + 32 = 0
12cosec²θ - 31cosecθ + 20 = 0
12cosec²θ - 16cosecθ - 15cosecθ + 20 = 0
4cosecθ(3cosecθ - 4) - 5(3cosecθ - 4) = 0
(4cosecθ - 5)(3cosecθ - 4) = 0
cosecθ = $$\frac{5}{4}$$, cosecθ = $$\frac{4}{3}$$

$$ = \frac{{\sqrt 7 \times 3}}{7}$$

Alternate:
Go through option A

12cot²θ - 31cosecθ + 32 = 0
$$\eqalign{ & 12 \times \frac{9}{{16}} - \frac{{31 \times 5}}{4} + 32 = 0 \cr & 0 = 0 \cr} $$

$$\eqalign{ & \frac{{{{\cos }^4}\theta - {{\sin }^4}\theta + 2{{\sin }^2}\theta + 3}}{{\left( {{\text{cosec}}\,\theta + \cot \theta + 1} \right)\left( {{\text{cosec}}\,\theta - \cot \theta + 1} \right) - 2}} \cr & = \frac{{\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) + 2{{\sin }^2}\theta + 3}}{{\left( {{\text{cosec}}\,\theta + 1 + \cot \theta } \right)\left( {{\text{cosec}}\,\theta + 1 - \cot \theta } \right) - 2}} \cr & = \frac{{\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) + 2{{\sin }^2}\theta + 3}}{{{{\left( {{\text{cosec}}\,\theta + 1} \right)}^2} - {{\cot }^2}\theta - 2}} \cr & = \frac{{{{\cos }^2}\theta + {{\sin }^2}\theta + 3}}{{{\text{cose}}{{\text{c}}^2}\theta + 1 + 2{\text{cosec}}\,\theta - {{\cot }^2}\theta - 2}} \cr & = \frac{4}{{2{\text{cosec}}\,\theta }} \cr & = 2\sin \theta \cr} $$

sin(A + B) = $$\frac{{\sqrt 3 }}{2}$$
sin(A + B) = sin60°
A + B = 60° . . . . . . . .(i)
tan(A - B) = $$\frac{1}{{\sqrt 3 }}$$
tan(A - B) = tan30°
A - B = 30° . . . . . . . . (ii)
Equation (i) adding equation (ii)
$$\eqalign{ & A + B = {60^ \circ } \cr & \underline {A - B = {{30}^ \circ }} \cr & 2A\,\,\,\,\,\,\,\,\, = {90^ \circ } \cr} $$
A = 45°
B = 15°
(2A + 3B)
= (2 × 45° + 3 × 15°)
= 90° + 45°
= 135°

$$\eqalign{ & \sin \theta = \frac{9}{{41}} \cr & \frac{P}{H} = \frac{9}{{41}},\,\,B = 40 \cr & \Rightarrow \cot \theta = \frac{B}{P} = \frac{{40}}{9} \cr} $$

$$\eqalign{ & \sin 10 - \frac{4}{3}{\sin ^3}10 \cr & = \frac{{3\sin 10 - 4{{\sin }^3}10}}{3} \cr & = \frac{1}{3}\sin 3 \times {10^ \circ }\,\,\,\,\,\left[ {\because \sin A = 3\sin A - 4{{\sin }^3}A} \right] \cr & = \frac{1}{3} \times \frac{1}{2} \cr & = \frac{1}{6} \cr} $$

$$\eqalign{ & \frac{{\sec A - \tan A}}{{\sec A + \tan A}} \cr & = \frac{{\sec A - \tan A}}{{\sec A + \tan A}} \times \frac{{\sec A - \tan A}}{{\sec A - \tan A}} \cr & = \frac{{{{\left( {\sec A - \tan A} \right)}^2}}}{{{{\sec }^2}A - {{\tan }^2}A}} \cr & = \frac{{{{\sec }^2}A + {{\tan }^2}A - 2.\sec A.\tan A}}{1} \cr & = 1 + {\tan ^2}A + {\tan ^2}A - 2.\sec A.\tan A \cr & = 1 + 2{\tan ^2}A - 2.\sec A.\tan A \cr} $$

$$\eqalign{ & A = {10^ \circ } \cr & \frac{{12\sin 3A + 5\cos \left( {5A - {5^ \circ }} \right)}}{{9\sin \frac{{9A}}{2} - 4\cos \left( {5A + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin 3 \times {{10}^ \circ } + 5\cos \left( {5 \times {{10}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{9 \times {{10}^ \circ }}}{2} - 4\cos \left( {5 \times {{10}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos \left( {{{50}^ \circ } - {5^ \circ }} \right)}}{{9\sin \frac{{{{90}^ \circ }}}{2} - 4\cos \left( {{{50}^ \circ } + {{10}^ \circ }} \right)}} \cr & = \frac{{12\sin {{30}^ \circ } + 5\cos {{45}^ \circ }}}{{9\sin {{45}^ \circ } - 4\cos {{60}^ \circ }}} \cr & = \frac{{12 \times \frac{1}{2} + 5 \times \frac{1}{{\sqrt 2 }}}}{{9 \times \frac{1}{{\sqrt 2 }} - 4 \times \frac{1}{2}}} \cr & = \frac{{6 + \frac{5}{{\sqrt 2 }}}}{{\frac{9}{{\sqrt 2 }} - 2}} \cr & = \frac{{6\sqrt 2 + 5}}{{9 - 2\sqrt 2 }} \cr} $$

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

cos(36° - A)cos(36° + A) + cos(54° - A)cos(54° + A)
= cos(36° - A)cos(36° + A) + sin(36° + A)sin(36° - A)
∵ cos(A - B) = cosA.cosB + sinA.sinB
= cos[36° - A - 36° - A]
= cos(-2A)
= cos2A

$$\eqalign{ & 3\left( {{{\cot }^2}\theta - {{\cos }^2}\theta } \right) = {\cos ^2}\theta \cr & 3{\cot ^2}\theta = {\cos ^2}\theta + 3{\cos ^2}\theta \cr & 3{\cot ^2}\theta = 4{\cos ^2}\theta \cr & \frac{3}{4} = {\sin ^2}\theta \cr & \sin \theta = \sin {60^ \circ } \cr & \theta = {60^ \circ } \cr & {\tan ^2}\theta + {\text{cose}}{{\text{c}}^2}\theta + {\sin ^2}\theta \cr & = {\tan ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{60^ \circ } + {\sin ^2}{60^ \circ } \cr & = 3 + \frac{4}{3} + \frac{3}{4} \cr & = \frac{{36 + 16 + 9}}{{12}} \cr & = \frac{{61}}{{12}} \cr} $$