Practice MCQ Questions and Answer on Trigonometry

Solution:

 ∵ P + Q + R = 60° By putting P = 0°, Q = 0° and R = 60° ⇒ cosQcosR(cosP - sinP) + sinQsinR(sinP - cosP) ⇒ 1 × cos60°(cos0° - sin0°) + sin0°.sin60°(sin0° - cos0°) ⇒ $$\frac{1}{2}$$ (1 - 0) + 0 ⇒ $$\frac{1}{2}$$

Solution:

 $$\eqalign{ & {\bf{Shortcut \,\, method:}} \cr & = \left( {1 + \cot \theta - \operatorname{cosec} \theta } \right)\left( {1 + \tan \theta + \sec \theta } \right) \cr & \left[ {put,\theta = {{45}^ \circ }} \right] \cr} $$   $$ = \left( {1 + \cot {{45}^ \circ } - \operatorname{cosec} {{45}^ \circ }} \right)$$     $$\left( {1 + \tan {{45}^ \circ } + \sec {{45}^ \circ }} \right)$$ $$\eqalign{ & = \left( {1 + 1 - \sqrt 2 } \right)\left( {1 + 1 + \sqrt 2 } \right) \cr & = \left( {2 - \sqrt 2 } \right)\left( {2 + \sqrt 2 } \right) \cr & = \left[ {{2^2} - {{\left( {\sqrt 2 } \right)}^2}} \right]\left[ {\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}} \right] \cr & = 4 - 2 \cr & = 2 \cr} $$

Solution:

 In these type of questions we can go through option- From option B $$\eqalign{ & \frac{{\left( {2\sin A} \right)\left( {1 + \sin A} \right)}}{{\left( {1 + \sin A} \right) + \cos A}} = 1 - \cos A + \sin A \cr & 2\sin A\left( {1 + \sin A} \right) = {\left( {1 + \sin A} \right)^2} - {\cos ^2}A \cr & 2\sin A\left( {1 + \sin A} \right) = 1 + \sin 2A + 2\sin A - 1 + {\sin ^2}A \cr & 2\sin A\left( {1 + \sin A} \right) = 2\sin A\left( {1 + \sin A} \right) \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put }}A = {90^ \circ } \cr & \frac{{\left( {2\sin {{90}^ \circ }} \right)\left( {1 + \sin {{90}^ \circ }} \right)}}{{1 + \sin {{90}^ \circ } + \cos {{90}^ \circ }}} = 2 \cr & {\text{Now from option B}} \cr & 1 + \sin A - \cos A \cr & = 1 + \sin {90^ \circ } - \cos {90^ \circ } \cr & = 1 + 1 - 0 \cr & = 2 \cr} $$

Solution:

 Let x = 8cos10°. cos20°. cos40° Multiply on both side by sin10° and applying formula (2sinθ. cosθ = sin2θ) ⇒ x sin10° = 4 × 2sin10° cos10°. cos20°. cos40° ⇒ x sin10° = 2 × 2sin20°.cos20°. cos40° ⇒ x sin10° = 2 × sin40°. cos40° ⇒ x sin10° = sin80° ⇒ x sin10° = sin(90° - 10°) ⇒ x sin10° = cos10° then, x = $$\frac{{\cos {{10}^ \circ }}}{{\sin {{10}^ \circ }}}$$ x = cot10°

Solution:

 (secA - cosA)2 + (cosecA - sinA)2 - (cotA - tanA)2 = (sec2A + cos2A - 2secA.cosA) + (coses2A + sin2A - 2cosecA.sinA) - (cot2A + tan2A - 2cotA.tanA) = sec2A - tan2A + cos2A + sin2A + coses2A - cot2A - 2 = 3 - 2 = 1 Alternate shortcut method: (secA - cosA)2 + (cosecA - sinA)2 - (cotA - tanA)2 Put θ = 45° = (sec45° - cos45°)2 + (cosec45° - sin45°)2 - (cot45° - tan45°)2 $$\eqalign{ & = {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} + {\left( {\sqrt 2 - \frac{1}{{\sqrt 2 }}} \right)^2} - {\left( {1 - 1} \right)^2} \cr & = \frac{1}{2} + \frac{1}{2} - 0 \cr & = 1 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \frac{{\cos \left( {\pi - A} \right).\cot \left( {\frac{\pi }{2} + A} \right)\cos \left( { - A} \right)}}{{\tan \left( {\pi + A} \right)\tan \left( {\frac{{3\pi }}{2} + A} \right)\sin \left( {2\pi - A} \right)}}\, \cr & = \frac{{\left( { - \cos A} \right) \times \left( { - \tan A} \right) \times \cos A}}{{\tan A \times \left( { - \cot A} \right) \times \left( { - \sin A} \right)}} \cr & = \frac{{{{\cos }^2}A}}{{\frac{{\cos A}}{{\sin A}} \times \sin A}} \cr & = \frac{{{{\cos }^2}A}}{{\cos A}} \cr & = \cos A \cr} $$

Solution:

 $$\eqalign{ & {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr & \Rightarrow {\text{co}}{{\text{s}}^4}\theta = 1 - {\cos ^2}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^4}\theta = {\sin ^2}\theta \cr & \Rightarrow {\cos ^2}\theta .{\cos ^2}\theta = {\sin ^2}\theta \cr & \Rightarrow {\cos ^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta = {\text{ta}}{{\text{n}}^2}\theta \cr & \Rightarrow {\text{ta}}{{\text{n}}^2}\theta + {\text{ta}}{{\text{n}}^4}\theta \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\theta + {\text{co}}{{\text{s}}^4}\theta = 1 \cr} $$

Solution:

 $$\eqalign{ & {\text{7}}{\sin ^2}\theta + 3{\text{co}}{{\text{s}}^2}\theta = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3\left( {{\text{1}} - {\text{si}}{{\text{n}}^2}\theta } \right) = 4 \cr & \Rightarrow {\text{7}}{\sin ^2}\theta + 3 - 3{\sin ^2}\theta = 4 \cr & \Rightarrow 4{\sin ^2}\theta = 1 \cr & \Rightarrow {\sin ^2}\theta = \frac{1}{4} \cr & \Rightarrow \sin \theta = \frac{1}{2} = {\text{sin 3}}{0^ \circ } \cr & \theta = {30^ \circ } = \frac{\pi }{6}\left[ {\because {\pi ^c} = {{180}^ \circ }} \right] \cr} $$

Solution:

 $$\eqalign{ & \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta - 3\cos \theta + 2}} = 1 \cr & {\sin ^2}\theta = {\cos ^2}\theta - 3\cos \theta + 2 \cr & 1 - {\cos ^2}\theta = {\cos ^2}\theta - 3\cos \theta + 2 \cr & 0 = 2{\cos ^2}\theta - 3\cos \theta + 1 \cr & 2{\cos ^2}\theta - 3\cos \theta + 1 = 0 \cr & 2{\cos ^2}\theta - 2\cos \theta - \cos \theta + 1 = 0 \cr & 2\cos \theta \left( {\cos \theta - 1} \right) - 1\left( {\cos \theta - 1} \right) = 0 \cr & \left( {2\cos \theta - 1} \right)\left( {\cos \theta - 1} \right) = 0 \cr & \cos \theta = \frac{1}{2} = \cos {60^ \circ } \cr & \cos \theta = 1 = \cos {0^ \circ } \cr & \Rightarrow \frac{{{{\tan }^2}\frac{\theta }{2} + {{\sin }^2}\frac{\theta }{2}}}{{\tan \theta + \sin \theta }} \cr & = \frac{{{{\tan }^2}\frac{{{{60}^ \circ }}}{2} + {{\sin }^2}\frac{{{{60}^ \circ }}}{2}}}{{\tan {{60}^ \circ } + \sin {{60}^ \circ }}} \cr & = \frac{{{{\tan }^2}{{30}^ \circ } + {{\sin }^2}{{30}^ \circ }}}{{\tan {{60}^ \circ } + \sin {{60}^ \circ }}} \cr & = \frac{{{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}}}{{\sqrt 3 + \frac{{\sqrt 3 }}{2}}} \cr & = \frac{{\frac{1}{3} + \frac{1}{4}}}{{\frac{{2\sqrt 3 + \sqrt 3 }}{2}}} \cr & = \frac{{\frac{{4 + 3}}{{12}}}}{{\frac{{3\sqrt 3 }}{2}}} \cr & = \frac{{7 \times 2}}{{12 \times 3\sqrt 3 }} \cr & = \frac{7}{{18\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{7\sqrt 3 }}{{54}} \cr} $$