Practice MCQ Questions and Answer on Trigonometry

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:

 $$\eqalign{ & \sec \theta + \tan \theta = p \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{p}{{\frac{1}{p}}} \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{{{p^2}}}{1} \cr & {\text{Apply componendo and dividendo}} \cr & \frac{{\sec \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{\sec \theta .\cos \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{{\text{cosec }}\theta }}{1} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & {\text{Apply again componendo and dividendo}} \cr & \frac{{{\text{cosec }}\theta + 1}}{{{\text{cosec }}\theta - 1}} = \frac{{{p^2}}}{1} \cr} $$

Solution:

 $$\eqalign{ & {\text{The value of }} \cr & {\text{3}}\left( {{{\sin }^4}\theta + {\text{co}}{{\text{s}}^4}\theta } \right) + 2\left( {{{\sin }^6}\theta + {\text{co}}{{\text{s}}^6}\theta } \right) + 12{\sin ^2}\theta .{\text{co}}{{\text{s}}^2}\theta \cr & {\text{Using }}\theta = {0^ \circ } \cr & \because \sin {0^ \circ } = {0^ \circ } \cr & \cos {0^ \circ } = 1 \cr & \Rightarrow 3\left( {0 + {1^4}} \right) + 2\left( {0 + {1^6}} \right) + 12 \times 0 \times 1 \cr & \Rightarrow 3 + 2 \cr & \Rightarrow 5 \cr} $$

Solution:

 $$\eqalign{ & m\left[ {\sin \theta + 2{{\cos }^2}\phi + 3\sin \theta + 4{{\cos }^2}\phi + \,.....\,18{{\cos }^2}\phi } \right] \cr & \theta = {30^ \circ },\,\,\phi = {45^ \circ } \cr & = m\left[ {\left( {\sin \theta + 3\sin \theta + 5\sin \theta + \,.....\,17\sin \theta } \right) + \left( {2{{\cos }^2}\phi + 4{{\cos }^2}\phi + \,.....\,18{{\cos }^2}\phi } \right)} \right] \cr & \Rightarrow {\text{Sum of odd number}} = {\left( 9 \right)^2}, \cr & {\text{Sum of even number}} = 9 \times 10 \cr & = m\left[ {81\sin \theta + 90{{\cos }^2}\phi } \right] \cr & = m\left[ {81 \times \sin 30 + 90 \times {{\cos }^2}45} \right] \cr & = m\left[ {81 \times \frac{1}{2} + 90 \times \frac{1}{2}} \right] \cr & = m \times \frac{{171}}{2} \cr & {\text{Now, we will check through the option,}} \cr & {\text{Putting }}m = 152 \cr & = 152 \times \frac{{171}}{2} \cr & = 4 \times 19 \times 19 \times 9 \cr & = {\left( {2 \times 19 \times 3} \right)^2} \cr & = {\text{ perfect square}}{\text{. Ans}}{\text{.}} \cr} $$

Solution:

 $$\eqalign{ & {\text{In }}\vartriangle {\text{ABC sin2}}{{\text{1}}^ \circ }{\text{ = }}\frac{x}{y} \cr & {\text{AB}} = x \cr & {\text{AC}} = y \cr & {\text{BC}} = \sqrt {{y^2} - {x^2}} \cr & \Rightarrow {\text{sec2}}{{\text{1}}^ \circ } - \sin {69^ \circ } \cr & \Rightarrow \frac{{{\text{AC}}}}{{{\text{BC}}}} - \frac{{{\text{BC}}}}{{{\text{AC}}}} \cr & \Rightarrow \frac{{{{\left( {{\text{AC}}} \right)}^2} - {{\left( {{\text{BC}}} \right)}^2}}}{{\left( {{\text{BC}}} \right)\left( {{\text{AC}}} \right)}} = \frac{{{y^2} - {{\left( {\sqrt {\left( {{y^2} - {x^2}} \right)} } \right)}^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr & \Rightarrow \frac{{{y^2} - {y^2} + {x^2}}}{{y\sqrt {{y^2} + {x^2}} }} = \frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr} $$

Solution:

 $$\eqalign{ & {a^2} + {b^2} + {c^2} = ab + bc + ca \cr & \Rightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 \cr & \Rightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0 \cr & \Rightarrow {a^2} + {b^2} - 2ab + {b^2} + {c^2} - 2bc + {c^2} + {a^2} - 2ca = 0 \cr & \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0 \cr & \therefore a = b = c \cr & \vartriangle {\text{ABC}} = {\text{equilateral }}\vartriangle \cr & \therefore \angle {\text{A}} = \angle {\text{B}} = \angle {\text{C}} = {60^ \circ } \cr & {\text{So, }}{\sin ^2}{\text{A}} + {\sin ^2}{\text{B}} + {\sin ^2}{\text{C}} \cr & \Rightarrow {\sin ^2}{60^ \circ } + {\sin ^2}{60^ \circ } + {\sin ^2}{60^ \circ } \cr & \Rightarrow 3{\sin ^2}{60^ \circ } \cr & \Rightarrow 3 \times {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \cr & \Rightarrow \frac{9}{4} \cr} $$

Solution:

 $$\eqalign{ & \frac{{2{{\tan }^2}{{30}^ \circ }}}{{1 - {{\tan }^2}{{30}^ \circ }}} + {\sec ^2}{45^ \circ } - {\sec ^2}{0^ \circ } = x\sec {60^ \circ } \cr & \Rightarrow \frac{{2 \times {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}{{1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}} + {\left( {\sqrt 2 } \right)^2} - 1 = x \times 2 \cr & \Rightarrow \frac{{2 \times \frac{1}{3}}}{{1 - \frac{1}{3}}} + 2 - 1 = 2x \cr & \Rightarrow \left( {\frac{2}{3} \times \frac{3}{2}} \right) + 2 - 1 = 2x \cr & \Rightarrow 2 = x \times 2 \cr & \Rightarrow x = 1 \cr} $$

Solution:

 $$\eqalign{ & \tan A - \tan B - \tan C = \tan A.\tan B.\tan C \cr & \tan A - \tan A.\tan B.\tan C = \tan B + \tan C \cr & \tan A = \frac{{\tan B + \tan C}}{{1 - \tan B.\tan C}} \cr & \tan A = \tan \left( {B + C} \right) \cr & {\text{On comparision,}} \cr & A = B + C \cr} $$

Solution:

 6sin4θ + 3cos4θ = 2 ⇒ 6sin4θ + 3(1 - sin2θ )2 = 2 ⇒ 6sin4θ + 3 + 3sin4θ - 6sin2θ = 2 ⇒ 9sin4θ - 6sin2θ + 1 = 0 ⇒ (3sin2θ -1)2 = 0 ⇒ 3sin2θ = 1 ⇒ sinθ = $$\frac{1}{{\sqrt 3 }} $$ Now, $${\left[ {7{{\operatorname{cosec} }^6}\theta + 8{{\sec }^6}\theta } \right]^{\frac{1}{3}}}$$ $$\eqalign{ & = {\left[ {7 \times {{(\sqrt 3 )}^6} + 8{{\left( {\frac{{\sqrt 3 }}{{\sqrt 2 }}} \right)}^6}} \right]^{\frac{1}{3}}} \cr & = {\left[ {7 \times 27 + 8 \times \frac{{27}}{8}} \right]^{\frac{1}{3}}} \cr & = {\left( {8 \times 27} \right)^{\frac{1}{3}}} \cr & = 6 \cr} $$

Solution:

 Given, cos27° = x tan63° = $$\frac{x}{{\sqrt {1 - {x^2}} }}$$