Practice MCQ Questions and Answer on Trigonometry

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:

 $$\eqalign{ & \tan 2\theta .\tan 3\theta = 1 \cr & \left( {2\theta + 3\theta } \right) = {90^ \circ } \cr & 5\theta = {90^ \circ } \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1}}} \right] \cr & \left[ {{\text{then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow \left( {{\text{2co}}{{\text{s}}^2}\frac{{5\theta }}{2} - 1} \right) \cr & \Rightarrow {\text{2co}}{{\text{s}}^2}\frac{{{{90}^ \circ }}}{2} - 1 \cr & \Rightarrow {\text{2co}}{{\text{s}}^2}{45^ \circ } - 1 \cr & \Rightarrow \frac{{\text{2}}}{2} - 1 \cr & \Rightarrow 0 \cr} $$

Solution:

 $$\eqalign{ & {\text{According to the question,}} \cr & {\text{Put A}} = {45^ \circ } \cr & \Rightarrow \frac{1}{2} \times \cot {45^ \circ } \cr & \left[ {\frac{{1 + \left( {\sec {{45}^ \circ } + \tan {{45}^ \circ }} \right)}}{{{\text{cosec }}{{45}^ \circ }\left( {\sec {{45}^ \circ } - \tan {{45}^ \circ }} \right)}}} \right] \cr & \Rightarrow \frac{1}{2}\left[ {\frac{{1 + {{\left( {\sqrt 2 - 1} \right)}^2}}}{{\sqrt 2 \times \left( {\sqrt 2 - 1} \right)}}} \right] \cr & \Rightarrow \frac{1}{2}\left[ {\frac{{1 + 2 + 1 - 2\sqrt 2 }}{{2 - \sqrt 2 }}} \right] \cr & \Rightarrow \frac{1}{2}\left[ {\frac{{4 - 2\sqrt 2 }}{{2 - \sqrt 2 }}} \right] \cr & \Rightarrow \frac{1}{2} \times 2\left[ {\frac{{2 - \sqrt 2 }}{{2 - \sqrt 2 }}} \right] \cr & \Rightarrow 1 \cr} $$

Solution:

 $${\text{cos}}\theta = \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$$ AC2 = (x2 + y2)2 - (x2 - y2)2         = x4 + y4 + 2x2y2 - x4 - y4 + 2x2y2         = 4x2y2 ⇒ AC = 2xy ⇒ cotθ = $$\frac{{{x^2} - {y^2}}}{{2xy}}$$

Solution:

 The value of, cos24° + cos55° + cos125° + cos204° + cos300° We know that, cos(180° $$ \pm $$ θ) = -cosθ ⇒ cos24° + cos55° + cos (180° - 55°) + cos (180° + 24°) + cos (360° - 60°) ⇒ cos24° + cos55° - cos55° - cos24° + cos60° ⇒ cos60° ⇒ $$\frac{1}{2}$$

Solution:

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

Solution:

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

Solution:

 A + B = C tan(A + B) = tanC $$\frac{{\tan {\text{A}} + \tan {\text{B}}}}{{1 - \tan {\text{A}}\tan {\text{B}}}} = \tan {\text{C}}$$ tanA + tanB = tanC - tanAtanBtanC tanAtanBtanC = tanC - tanA - tanB

Solution:

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

Solution:

 A + B = C tan(A + B) = tanC $$\frac{{\tan {\text{A}} + \tan {\text{B}}}}{{1 - \tan {\text{A}}\tan {\text{B}}}} = \tan {\text{C}}$$ tanA + tanB = tanC - tanAtanBtanC tanAtanBtanC = tanC - tanA - tanB