Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & 3{\sec ^2}\theta + \tan \theta - 7 = 0 \cr & \Rightarrow 3\left( {1 + {{\tan }^2}\theta } \right) + \tan \theta - 7 = 0 \cr & \Rightarrow 3{\tan ^2}\theta + \tan \theta - 4 = 0 \cr & \Rightarrow 3{\tan ^2}\theta + 4\tan \theta - 3\tan \theta - 4 = 0 \cr & \Rightarrow \tan \theta \left( {3\tan \theta + 4} \right) - 1\left( {3\tan \theta + 4} \right) = 0 \cr & \Rightarrow \left( {3\tan \theta + 4} \right)\left( {\tan \theta - 1} \right) = 0 \cr & \Rightarrow \tan \theta = 1\,\,\,\,\,\,\,\,\,\therefore \theta = {45^ \circ } \cr & \therefore \,\frac{{2\sin \theta + 3\cos \theta }}{{{\text{cosec}}\,\theta + \sec \theta }} \cr & = \frac{{2\left( {\frac{1}{{\sqrt 2 }}} \right) + 3\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\sqrt 2 + \sqrt 2 }} \cr & = \frac{{\frac{5}{{\sqrt 2 }}}}{{2\sqrt 2 }} \cr & = \frac{5}{4} \cr} $$

sin(90 - x)cos[π - (x - y)] + cos(90 - x)sin[π - (y - x)]
⇒ cosx{-cos(x - y)} + sinx.sin(y - x)
⇒ -cosx.cos(x - y) + sinx.sin(y - x)
⇒ -cosx(cosx.cosy + sinx.siny) + sinx(siny.cosx - cosy.sinx)
⇒ -cos²x.cosy - cosx.sinx.siny + sinx.siny.cosx - cosy.sin²x
⇒ -cosy(cos²x + sin²x)
⇒ -cosy

$$\eqalign{ & \frac{{3\left( {{\text{cose}}{{\text{c}}^2}{{26}^ \circ } - {{\tan }^2}{{64}^ \circ }} \right) + \left( {{{\cot }^2}{{42}^ \circ } - {{\sec }^2}{{48}^ \circ }} \right)}}{{\cot \left( {{{22}^ \circ } - \theta } \right) - {\text{cose}}{{\text{c}}^2}\left( {{{62}^ \circ } + \theta } \right) - \tan \left( {\theta + {{68}^ \circ }} \right) + {{\tan }^2}\left( {{{28}^ \circ } - \theta } \right)}} \cr & = \frac{{3\left( {{\text{cose}}{{\text{c}}^2}{{26}^ \circ } - {{\cot }^2}{{26}^ \circ }} \right) + \left( {{{\cot }^2}{{42}^ \circ } - {\text{cose}}{{\text{c}}^2}{{42}^ \circ }} \right)}}{{\cot \left( {{{22}^ \circ } - \theta } \right) - {\text{cose}}{{\text{c}}^2}\left( {{{62}^ \circ } + \theta } \right) - \cot \left( {{{22}^ \circ } + \theta } \right) + {{\cot }^2}\left( {{{62}^ \circ } + \theta } \right)}} \cr & = \frac{{3 \times 1 + \left( { - 1} \right)}}{{ - 1}} \cr & = - 2 \cr} $$

$$\eqalign{ & \frac{{\sin \left( {A + B} \right)}}{{\sin A.\cos B}} \cr & = \frac{{\sin A.\cos B + \cos A.\sin B}}{{\sin A.\cos B}} \cr & = 1 + \frac{{\cos A.\sin B}}{{\sin A.\cos B}} \cr & = 1 + \cot A.\tan B \cr} $$

sinθ + cosθ = √5cosθ
sinθ = cosθ(√5 - 1)
$$\cot \theta = \frac{1}{{\sqrt 5 - 1}} \times \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} = \frac{{\sqrt 5 + 1}}{4}$$

$$\eqalign{ & 2\frac{{{{\cos }^2}x - {{\sec }^2}x}}{{{{\tan }^2}x}} = a + b\cos 2x \cr & \Rightarrow 2\frac{{{{\cos }^2}x - \frac{1}{{{{\cos }^2}x}}}}{{\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}} = a + b\cos 2x \cr & \Rightarrow 2\frac{{{{\cos }^4}x - 1}}{{{{\cos }^2}x}} \times \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = a + b\cos 2x \cr & \Rightarrow 2\frac{{\left( {{{\cos }^2}x - 1} \right)\left( {{{\cos }^2}x + 1} \right)}}{{\left( {1 - {{\cos }^2}x} \right)}} = a + b\cos 2x \cr & \Rightarrow \frac{{ - 2\left( {1 - {{\cos }^2}x} \right)\left( {{{\cos }^2}x + 1} \right)}}{{\left( {1 - {{\cos }^2}x} \right)}} = a + b\cos 2x \cr & \Rightarrow - 2{\cos ^2}x - 2 = a + b\cos 2x \cr & \Rightarrow - 2 + 1 - 1 - 2{\cos ^2}x = a + b\cos 2x \cr & \Rightarrow - 3 - \left( {2{{\cos }^2}x - 1} \right) = a + b\cos 2x \cr & \Rightarrow - 3 - \cos 2x = a + b{\cos ^2}x \cr & a = - 3,\,\,b = - 1 \cr} $$

$$\eqalign{ & \frac{{\cos {{50}^ \circ }}}{{\sin {{40}^ \circ }}} + \frac{{3{\text{cosec}}\,{\text{8}}{0^ \circ }}}{{\sec {{10}^ \circ }}} - 2\cos {50^ \circ } \cdot {\text{cosec}}\,{40^ \circ }{\text{ angle of sum}} = {90^ \circ } \cr & {\text{then,}} \cr & = \frac{{\sin {{40}^ \circ }}}{{\sin {{40}^ \circ }}} + \frac{{3{\text{cosec}}\,{\text{5}}{0^ \circ }}}{{{\text{cosec}}\,{\text{8}}{0^ \circ }}} - 2\sin {40^ \circ } \cdot {\text{cosec}}\,{10^ \circ } \cr & = 1 + 3 - 2 \cr & = 2 \cr} $$

$$\eqalign{ & 3\sin \theta = 2{\cos ^2}\theta \cr & {\text{Let }}\theta = {30^ \circ } \cr & 3 \times \frac{1}{2} = 2 \times \frac{{{{\left( {\sqrt 3 } \right)}^2}}}{4} \cr & \frac{3}{2} = \frac{3}{2} \cr & {\tan ^2}\theta + {\sec ^2}\theta - {\text{cose}}{{\text{c}}^2}\theta \cr & = {\tan ^2}{30^ \circ } + {\sec ^2}{30^ \circ } - {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr & = \frac{1}{3} + \frac{4}{3} - 4 \cr & = \frac{5}{3} - 4 \cr & = - \frac{7}{3} \cr} $$

cos(A - α) = p, sin(A - β) = q
Here, two equation, and four variable,
So, α = β = 0,
cosA = p, sinA = q
Now, cos²(α - β) + 2pq sin(α - β)
cos²0 + 2pq × sin0
1 + 0 = 1
By option, (C) = p² + q²
= cos²A + sin²A
= 1

$$\eqalign{ & \frac{{{{\cos }^2}\theta }}{{{{\cot }^2}\theta - {{\cos }^2}\theta }} = 3 \cr & \frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta \left( {\frac{{1 - {{\sin }^2}\theta }}{{{{\sin }^2}\theta }}} \right)}} = 3 \cr & \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = 3 \cr & \tan \theta = \sqrt 3 \cr & \theta = {60^ \circ } \cr & \cot \theta + {\text{cosec}}\,\theta \cr & = \cot {60^ \circ } + {\text{cosec}}\,{60^ \circ } \cr & = \frac{1}{{\sqrt 3 }} + \frac{2}{{\sqrt 3 }} \cr & = \frac{3}{{\sqrt 3 }} \cr & = \sqrt 3 \cr} $$