Practice MCQ Questions and Answer on Trigonometry

Solution:

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

Solution:

 $$\eqalign{ & 1 + 2{\tan ^2}\theta + 2\sin \theta {\sec ^2}\theta = \frac{a}{b} \cr & 1 + {\tan ^2}\theta + {\tan ^2}\theta + 2\sin \theta \frac{1}{{{{\cos }^2}\theta }} = \frac{a}{b} \cr & {\sec ^2}\theta + {\tan ^2}\theta + 2\tan \theta \sec \theta = \frac{a}{b} \cr & {\left( {\sec \theta + \tan \theta } \right)^2} = \frac{a}{b} \cr & {\left( {\frac{1}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }}} \right)^2} = \frac{a}{b} \cr & {\left( {\frac{{1 + \sin \theta }}{{\cos \theta }}} \right)^2} = \frac{a}{b} \cr & \frac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{1 + {{\sin }^2}\theta }} = \frac{a}{b} \cr & \frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{a}{b} \cr & {\text{Use componendo and dividendo}} \cr & \frac{1}{{\sin \theta }} = {\text{cosec}}\,\theta = \frac{{a + b}}{{a - b}} \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \sin \left( {A + B} \right) = \cos \left( {A + B} \right) \cr & \frac{{\sin \left( {A + B} \right)}}{{\cos \left( {A + B} \right)}} = 1 \cr & \tan \left( {A + B} \right) = 1 \cr & \tan \left( {A + B} \right) = \tan {45^ \circ } \cr & A + B = {45^ \circ } \cr & A = {45^ \circ } - B \cr & \tan A = \tan {45^ \circ } - \tan B \cr & \tan A = \frac{{\tan {{45}^ \circ } - \tan B}}{{1 + \tan {{45}^ \circ }\tan B}} \cr & \tan A = \frac{{1 - \tan B}}{{1 + \tan B}} \cr} $$

Solution:

 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

Solution:

 $$\eqalign{ & 5\sin \theta - 4\cos \theta = 0 \cr & 5\sin \theta = 4\cos \theta \cr & \frac{{\sin \theta }}{{\cos \theta }} = \frac{4}{5} \cr & \tan \theta = \frac{4}{5} \cr & \frac{{5\sin \theta - 2\cos \theta }}{{5\sin \theta + 3\cos \theta }} \cr & = \frac{{5\tan \theta - 2}}{{5\tan \theta + 3}} \cr & = \frac{{5 \times \frac{4}{5} - 2}}{{5 \times \frac{4}{5} + 2}} \cr & = \frac{2}{7} \cr} $$

Solution:

 $$\eqalign{ & {\text{4}}\angle {\text{A}} = {\text{3}}\angle {\text{B}} = {\text{12}}\angle {\text{C}} \cr & {\text{A}}:{\text{B}}:{\text{C}} = \frac{1}{4}:\frac{1}{3}:\frac{1}{{12}} \cr & {\text{A}}:{\text{B}}:{\text{C}} = 3:4:1 \cr & {\text{Now,}} \cr & 3x + 4x + x = {180^ \circ } \cr & 8x = {180^ \circ } \cr & x = \frac{{{{180}^ \circ }}}{8} \cr & \angle {\text{A}} = 3x = \frac{{{{180}^ \circ }}}{8} \times 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {67.5^ \circ } \cr} $$

Solution:

 $$\eqalign{ & x = a{\text{ }}\sec \theta .\cos \phi \cr & y = b{\text{ }}\sec \theta .sin\phi \cr & z = c{\text{ tan}}\theta \cr & \frac{x}{a} = \sec \theta .\cos \phi \cr & \frac{y}{b} = \sec \theta .sin\phi \cr & \frac{z}{c} = {\text{tan}}\theta \cr & \therefore \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} \cr & \Rightarrow {\sec ^2}\theta .{\cos ^2}\phi + {\sec ^2}\theta .si{n^2}\phi - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta \left( {{{\cos }^2}\phi + si{n^2}\phi } \right) - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow 1 \cr} $$

Solution:

 $$\eqalign{ & {\text{cose}}{{\text{c}}^4}A\left( {1 - {{\cos }^4}A} \right) - 2{\cot ^2}A - 1 \cr & {\text{Here we take }}A = {45^ \circ } \cr & = {\text{cose}}{{\text{c}}^4}{45^ \circ }\left( {1 - {{\cos }^4}{{45}^ \circ }} \right) - 2{\cot ^2}{45^ \circ } - 1 \cr & = {\left( {\sqrt 2 } \right)^4}\left\{ {1 - {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^4}} \right\} - 2\left( 1 \right) - 1 \cr & = 4\left( {1 - \frac{1}{4}} \right) - 2 - 1 \cr & = 4\left( {\frac{3}{4}} \right) - 2 - 1 \cr & = 3 - 2 - 1 \cr & = 0 \cr} $$

Solution:

 $$\eqalign{ & \frac{{{{\tan }^6}\theta - {{\sec }^6}\theta + 3{{\sec }^2}\theta \,{{\tan }^2}\theta }}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{{{\left( {{{\tan }^2}\theta } \right)}^3} - {{\left( {{{\sec }^2}\theta } \right)}^3} + 3{{\sec }^2}\theta \,{{\tan }^2}\theta }}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & \left[ {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)} \right] \cr & = \frac{{\left( {{{\tan }^2}\theta - {{\sec }^2}\theta } \right)\left( {{{\tan }^4}\theta + {{\sec }^4}\theta + {{\tan }^2}\theta {{\sec }^2}\theta + 3{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\tan }^4}\theta + {{\sec }^4}\theta + {{\tan }^2}\theta {{\sec }^2}\theta - 3{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\tan }^4}\theta + {{\sec }^4}\theta - 2{{\sec }^2}\theta {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)}}{{{{\tan }^2}\theta + {{\cot }^2}\theta + 2}} \cr & = \frac{{ - 1}}{{{{\left( {\tan \theta + \cot \theta } \right)}^2}}} \cr & = \frac{{ - 1}}{{{{\left( {\frac{{\sin \theta }}{{\cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}} \right)}^2}}} \cr & = \frac{{ - 1}}{{{{\left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta \sin \theta }}} \right)}^2}}} \cr & = - {\left( {\cos \theta \sin \theta } \right)^2} \cr & = - {\cos ^2}\theta {\sin ^2}\theta \cr} $$