Practice MCQ Questions and Answer on Trigonometry

Solution:

 sin31° = $$\frac{x}{y}{\text{,}}$$ ∴ sec31° - sin59° $$\eqalign{ & = \frac{y}{{\sqrt {{y^2} - {x^2}} }} - \frac{{\sqrt {{y^2} - {x^2}} }}{y} \cr & = \frac{{{y^2} - ({y^2} - {x^2})}}{{y\sqrt {{y^2} - {x^2}} }} \cr & = \frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr} $$

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:

 $$\eqalign{ & \sin \theta \times \cos \theta = \frac{1}{2} \cr & {\text{Multiply by 2 both side}} \cr & 2\sin \theta \times \cos \theta = 1 \cr & \sin 2\theta = 1 \cr & \sin 2\theta = \sin {90^ \circ } \cr & 2\theta = {90^ \circ } \cr & \theta = {45^ \circ } \cr & {\text{So,}}\sin \theta - {\text{cos}}\theta \cr & = \sin {45^ \circ } - \cos {45^ \circ } \cr & = \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} \cr & = 0{\text{ }} \cr} $$

Solution:

 $$\eqalign{ & {\text{co}}{{\text{s}}^4}\theta - {\sin ^4}\theta = \frac{2}{3} \cr & \left[ {{a^4} - {b^4} = \left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right)} \right] \cr & \Rightarrow \left( {{\text{co}}{{\text{s}}^2}\theta + {{\sin }^2}\theta } \right)\left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right) = \frac{2}{3} \cr & \left[ {{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \right] \cr & \Rightarrow 1 \times \left( {{\text{co}}{{\text{s}}^2}\theta - {{\sin }^2}\theta } \right) = \frac{2}{3} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta - \left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right) = \frac{2}{3} \cr & \left[ {{{\sin }^2}\theta = 1 - {\text{co}}{{\text{s}}^2}\theta } \right] \cr & \Rightarrow 2{\text{co}}{{\text{s}}^2}\theta - 1 = \frac{2}{3} \cr} $$

Solution:

 $$\eqalign{ & \cos \left( {A - B} \right) = \frac{{\sqrt 3 }}{2} \cr & \cos \left( {A - B} \right) = \cos {30^ \circ } \cr & A - B = {30^ \circ }........\left( {\text{i}} \right) \cr & \sec A = 2 \cr & \cos A = \frac{1}{2} = \cos {60^ \circ } \cr & A = {60^ \circ }........\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & A = {60^ \circ }\,\& \,B = {30^ \circ } \cr} $$

Solution:

 $$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & 4{\sin ^2}\theta = 1 \cr & {\sin ^2}\theta = \frac{1}{4} \cr & {\text{sin }}\theta = \frac{1}{2} \cr & \sin \theta = \sin {30^ \circ } \cr & \theta = {30^ \circ } \cr & \sec \theta + \operatorname{cosec} \theta \cr & = \sec {30^ \circ } + \operatorname{cosec} {30^ \circ } \cr & = \frac{2}{{\sqrt 3 }} + 2 \cr} $$

Solution:

 3cos80°.cosec10° + 2cos59°.cosec31° [If A + B = 90° then, cosA.cosecB = 1] (3 × 1 + 2 × 1 = 5) (the value of sinθ is always between -1 to +1)   $$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin {{10}^ \circ }}}{\text{ + 2cos5}}{{\text{9}}^ \circ }\frac{1}{{{\text{sin 3}}{{\text{1}}^ \circ }}}$$   $$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin \left( {{{90}^ \circ } - {{80}^ \circ }} \right)}}{\text{ + }}$$     $${\text{2cos5}}{{\text{9}}^ \circ }$$  $$\frac{1}{{{\text{sin }}{{\left( {{{90}^ \circ } - {{59}^ \circ }} \right)}^ \circ }}}$$ $$\eqalign{ & \Rightarrow 3 + 2 \cr & \Rightarrow 5 \cr} $$

Solution:

 $$\eqalign{ & \pi \sin \theta = 1\,.....(i) \cr & \pi \cos \theta = 1\,.....(ii) \cr & {\text{Divide eq}}{\text{. (i) from (ii)}} \cr & \frac{{\pi \sin \theta }}{{\pi \cos \theta }} = \frac{1}{1} \cr & \tan \theta = 1 \cr & \tan \theta = \tan {45^ \circ } \cr & \theta = {45^ \circ } \cr & \therefore \sqrt 3 \tan \left( {\frac{2}{3}\theta } \right) + 1 \cr & = \sqrt 3 \tan \left( {\frac{2}{3} \times {{45}^ \circ }} \right) + 1 \cr & = \sqrt 3 {\text{ tan}}{30^ \circ } + 1 \cr & = \sqrt 3 \times \frac{1}{{\sqrt 3 }} + 1 \cr & = 2 \cr} $$

Solution:

 $$\eqalign{ & \sin A = \frac{5}{{13}} = \frac{P}{H};\,B = 12 \cr & \left[ {5,\,12,\,13\,{\text{are triplet}}} \right] \cr & 7\cot B = 24 \cr & \cot B = \frac{{24}}{7} = \frac{B}{P};\,H = 25 \cr & \left[ {7,\,24,\,25\,{\text{are triplet}}} \right] \cr & \Rightarrow \left( {\sec A\cos B} \right)\left( {{\text{cosec}}\,B\tan A} \right) \cr & = \left( {\frac{{13}}{{12}} \times \frac{{24}}{{25}}} \right)\left( {\frac{{25}}{7} \times \frac{5}{{12}}} \right) \cr & = \frac{{26}}{{25}} \times \left( {\frac{{25}}{7} \times \frac{5}{{12}}} \right) \cr & = \frac{{65}}{{42}} \cr} $$

Solution:

 $$\eqalign{ & \left( {\frac{{\sin \theta + \sin \phi }}{{\cos \theta + \cos \phi }} + \frac{{\cos \theta \cos \phi }}{{\sin\theta - \sin\phi }}} \right) \cr & {\text{Put }}\theta = {90^ \circ } \cr & \phi = {0^ \circ } \cr & \therefore \left( {\frac{{\sin90^ \circ + \sin0^ \circ }}{{\cos90^ \circ + \cos0^ \circ }} + \frac{{\cos90^ \circ - \cos0^ \circ }}{{\sin90^ \circ - \sin0^ \circ }}} \right) \cr & \Rightarrow \left( {\frac{{1 + 0}}{{0 + 1}} + \frac{{0 - 1}}{{1 - 0}}} \right) \cr & \Rightarrow \left( {1 - 1} \right) \cr & \Rightarrow 0 \cr} $$