Trigonometry MCQ Questions and Solutions with Explanations | Aptitude

$$\eqalign{ & \frac{1}{{\tan \theta }} + \tan \theta \cr & {\text{put }}\theta = {45^ \circ } \cr & \frac{1}{1} + 1 = 2 \cr & {\text{In option}} \cr & \Rightarrow \left( {\text{A}} \right)\frac{{{\text{cosec}}\,\theta }}{{\sec \theta }} = \frac{{{\text{cosec}}\,{{45}^ \circ }}}{{\sec {{45}^ \circ }}} = \frac{{\sqrt 2 }}{{\sqrt 2 }} = 1 \cr & \Rightarrow \left( {\text{B}} \right)\sec \theta \times {\text{cosec}}\,\theta \cr & = \sec {45^ \circ } \times {\text{cosec}}\,{45^ \circ } \cr & = \sqrt 2 \times \sqrt 2 = 2 \cr & \Rightarrow \left( {\text{C}} \right)\,1 \cr & \Rightarrow \left( {\text{D}} \right){\tan ^2}\theta = {\tan ^2}{45^ \circ } = 1 \cr} $$

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

$$\eqalign{ & \cos \theta = \frac{{2p}}{{1 + {p^2}}} \cr & A{B^2} = {\left( {1 + {p^2}} \right)^2} - {\left( {2p} \right)^2} \cr & AB = 1 - {p^2} \cr & \tan \theta = \frac{{1 - {p^2}}}{{2p}} \cr} $$

We need to find which option makes the equation a² = b² + (ab)² true.
Remember, we are dealing with trigonometric functions: sin(x), cos(x), tan(x), and cot(x).
Let's look at each option:
Option A: a = sin(x), b = cot(x)
This means our equation becomes: sin²(x) = cot²(x) + (sin(x) * cot(x))²
Remember that cot(x) = cos(x) / sin(x). So, we can rewrite the equation.
sin²(x) = (cos²(x) / sin²(x)) + (sin(x) * (cos(x) / sin(x)))²
sin²(x) = (cos²(x) / sin²(x)) + cos²(x)
This looks complicated, and it's unlikely to simplify easily to something that's always true.
Option B: a = cos(x), b = tan(x)
This means our equation becomes: cos²(x) = tan²(x) + (cos(x) * tan(x))²
Remember that tan(x) = sin(x) / cos(x). Substitute that in.
cos²(x) = (sin²(x) / cos²(x)) + (cos(x) * (sin(x) / cos(x)))²
cos²(x) = (sin²(x) / cos²(x)) + sin²(x)
Again, this doesn't look like it will easily simplify to a true statement.
Option C: a = cot(x), b = cos(x)
This means our equation becomes: cot²(x) = cos²(x) + (cot(x) * cos(x))²
Substitute cot(x) = cos(x) / sin(x).
(cos²(x) / sin²(x)) = cos²(x) + ((cos(x) / sin(x)) * cos(x))²
(cos²(x) / sin²(x)) = cos²(x) + (cos²(x) / sin²(x)) * cos²(x)
(cos²(x) / sin²(x)) = cos²(x) + (cos⁴(x) / sin²(x))
Let's try to manipulate this a bit. Multiply both sides by sin²(x):
cos²(x) = cos²(x)sin²(x) + cos⁴(x)
cos²(x) = cos²(x)(sin²(x) + cos²(x))
Remember the fundamental trigonometric identity: sin²(x) + cos²(x) = 1
cos²(x) = cos²(x) * 1
cos²(x) = cos²(x). This is always true!
Option D: a = sin(x), b = tan(x)
This means our equation becomes: sin²(x) = tan²(x) + (sin(x) * tan(x))²
Substitute tan(x) = sin(x) / cos(x).
sin²(x) = (sin²(x) / cos²(x)) + (sin(x) * (sin(x) / cos(x)))²
sin²(x) = (sin²(x) / cos²(x)) + (sin⁴(x) / cos²(x))
This also doesn't look like it will simplify easily.
Therefore, Option C is the correct answer. It's the only one that simplifies to a true identity.

tan240°
= tan(180° + 60°)
= tan(180° + θ)
= tanθ
= tan60°
=√3

$$\sin \theta = \frac{a}{{\sqrt {{a^2} + {b^2}} }}$$

$$\eqalign{ & \sec \theta + \tan \theta \cr & = \frac{{\sqrt {{a^2} + {b^2}} }}{b} + \frac{a}{b} \cr & = \frac{{\sqrt {{a^2} + {b^2}} + a}}{b} \cr} $$

cos²θ - sin²θ = $$\frac{1}{2}$$
cos2θ = $$\frac{1}{2}$$
cos2θ = cos60°
2θ = 60°
θ = 30
tan²2θ + sin²3θ
= tan²60° + sin²90°
= 3 + 1
= 4

tan(θ - 14π)
= -tan(14π - θ)
= -(-tanθ)
= tanθ

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

$$\eqalign{ & {\sec ^2}{28^ \circ } - {\cot ^2}{62^ \circ } + {\sin ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr & = {\sec ^2}{28^ \circ } - {\tan ^2}{28^ \circ } + {\sin ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{30^ \circ } \cr & = 1 + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + {\left( 2 \right)^2} \cr & = 1 + \frac{3}{4} + 4 \cr & = \frac{{23}}{4} \cr} $$