Trigonometric Ratio and Identities MCQ Questions & Answers in Trigonometry | Maths

$$\eqalign{ & {\text{Given, }}A = \left( {\cos {{12}^ \circ } - \cos {{36}^ \circ }} \right)\left( {\sin {{96}^ \circ } + \sin {{24}^ \circ }} \right) \cr & B = \left( {\sin {{60}^ \circ } - \sin {{12}^ \circ }} \right)\left( {\cos {{48}^ \circ } - \cos {{72}^ \circ }} \right) \cr & \frac{A}{B} = \frac{{\left[ { - 2\sin {{24}^ \circ }\sin {{12}^ \circ }} \right]\left[ {2\sin {{60}^ \circ }\cos {{36}^ \circ }} \right]}}{{\left[ {2\cos {{36}^ \circ }\sin {{24}^ \circ }} \right]\left[ { - 2\sin {{60}^ \circ }\sin {{12}^ \circ }} \right]}} \cr & \Rightarrow \frac{A}{B} = 1 \cr} $$

Value $$ = 2{\cos ^2}\theta - 1 + \cos \theta = - 1 + 2\left( {{{\cos }^2}\theta + \frac{1}{2}\cos \theta + \frac{1}{{16}}} \right) - \frac{1}{8}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \frac{9}{8} + 2{\left( {\cos \theta + \frac{1}{4}} \right)^2} \geqslant - \frac{9}{8}.$$
So, the minimum value $$ = - \frac{9}{8}.$$

$$\eqalign{ & {x^2} - 2x + 4 = - 3\cos \left( {ax + b} \right) \cr & \Rightarrow \,\,{\left( {x - 1} \right)^2} + 3 = - 3\cos \left( {ax + b} \right). \cr} $$
$${\text{As }} - 1 \leqslant \cos \left( {ax + b} \right) \leqslant 1\,\,{\text{and }}{\left( {x - 1} \right)^2} \geqslant 0,$$         the above is possible only if $$\cos\left( {ax + b} \right) = - 1\,\,{\text{when }}x = 1.\,{\text{So, }}a + b = \pi ,3\pi ,5\pi ,\,{\text{e}}{\text{.t}}{\text{.c}}{\text{., and }}3\pi > 6.$$

Use $$\cos \alpha + \cos \left( {\alpha + \beta } \right) + \cos \left( {\alpha + 2\beta } \right) + ..... + \cos \left( {\alpha + \overline {n - 1} \beta } \right)$$
$$\,\,\,\,\, = \frac{{\sin \frac{{n\beta }}{2}}}{{\sin \frac{\beta }{2}}}\cos \frac{{2\alpha + \left( {n - 1} \right)\beta }}{2}.$$
Here, $$\alpha = \frac{\pi }{{11}},\beta = \frac{{2\pi }}{{11}},n = 5.$$

The given expression
$$\eqalign{ & = \tan A + \left[ {\frac{{\sqrt 3 + \tan A}}{{1 - \sqrt 3 \tan A}}} \right] - \left[ {\frac{{\sqrt 3 - \tan A}}{{1 + \sqrt 3 \tan A}}} \right] \cr & = \tan A + \left[ {\frac{{8\tan A}}{{1 - 3\,{{\tan }^2}A}}} \right] = \frac{{9\tan A - 3\,{{\tan }^3}\,3}}{{1 - 3\,{{\tan }^2}A}} \cr & = 3 \cdot \frac{{\left( {3\tan A - {{\tan }^3}A} \right)}}{{1 - 3\,{{\tan }^2}A}} = 3\tan 3A \cr} $$

$$\eqalign{ & \frac{{\cot {{224}^ \circ } - \cot {{134}^ \circ }}}{{\cot {{226}^ \circ } + \cot {{316}^ \circ }}} \cr & = \frac{{\cot \left( {{{180}^ \circ } + {{44}^ \circ }} \right) - \cot \left( {{{180}^ \circ } - {{46}^ \circ }} \right)}}{{\cot \left( {{{180}^ \circ} + {{46}^ \circ }} \right) + \cot {{\left( {{{270}^ \circ } + {{46}^ \circ}} \right)} }}} \cr & = \frac{{\cot {{44}^ \circ } + \cot {{46}^ \circ }}}{{\cot {{46}^ \circ } - \tan {{46}^ \circ }}} = \frac{{\tan {{46}^ \circ } + \tan {{44}^ \circ }}}{{\tan {{44}^ \circ } - \tan {{46}^ \circ }}} \cr & = \frac{{\sin \left( {{{46}^ \circ } + {{44}^ \circ }} \right)}}{{\sin \left( {{{44}^ \circ } - {{46}^ \circ }} \right)}} = - \,{\text{cosec }}{{\text{2}}^ \circ } \cr} $$

$$\eqalign{ & \frac{{1 + \sin \alpha - \cos \alpha }}{{1 + \sin \alpha }} = \frac{{{{\left( {1 + \sin \alpha } \right)}^2} - {{\cos }^2}\alpha }}{{\left( {1 + \sin \alpha } \right)\left( {1 + \sin \alpha + \cos \alpha } \right)}} \cr & \frac{{1 + \sin \alpha - \cos \alpha }}{{1 + \sin \alpha }} = \frac{{2\sin \alpha + 2{{\sin }^2}\alpha }}{{\left( {1 + \sin \alpha } \right)\left( {1 + \sin \alpha + \cos \alpha } \right)}} \cr & \frac{{1 + \sin \alpha - \cos \alpha }}{{1 + \sin \alpha }} = \frac{{2\sin \alpha }}{{1 + \sin \alpha + \cos \alpha }} = \lambda . \cr} $$

$$\eqalign{ & {u^2} = {a^2} + {b^2} + 2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right)} \,\,\,......\left( 1 \right) \cr & {\text{Now}}\,\left( {{a^4} + {b^4}} \right){\cos ^2}\theta {\sin ^2}\theta + {a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right) \cr & = \left( {{a^4} + {b^4}} \right){\cos ^2}\theta {\sin ^2}\theta + {a^2}{b^2}\left( {1 - 2{{\cos }^2}\theta {{\sin }^2}\theta } \right) \cr & = \left( {{a^4} + {b^4} - 2{a^2}{b^2}} \right){\cos ^2}\theta {\sin ^2}\theta + {a^2}{b^2} \cr & = {\left( {{a^2} - {b^2}} \right)^2}.\frac{{{{\sin }^2}2\theta }}{4} + {a^2}{b^2}\,\,\,.....\left( 2 \right) \cr & \because \,\,0 \leqslant {\sin ^2}2\theta \leqslant 1 \cr & \Rightarrow \,\,0 \leqslant {\left( {{a^2} - {b^2}} \right)^2}\frac{{{{\sin }^2}2\theta }}{4} \leqslant \frac{{{{\left( {{a^2} - {b^2}} \right)}^2}}}{4} \cr & \Rightarrow \,\,{a^2}{b^2} \leqslant {\left( {{a^2} - {b^2}} \right)^2}\frac{{{{\sin }^2}2\theta }}{4} + {a^2}{b^2} \leqslant {\left( {{a^2} - {b^2}} \right)^2}.\frac{1}{4} + {a^2}{b^2}\,\,\,\,\,.....\left( 3 \right) \cr} $$
∴ from (1), (2) and (3)
Minimum value of $${u^2} = {a^2} + {b^2} + 2\sqrt {{a^2}{b^2}} = {\left( {a + b} \right)^2}$$
Maximum value of $${u^2}$$
$$\eqalign{ & = {a^2} + {b^2} + 2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}.\frac{1}{4} + {a^2}{b^2}} \cr & = {a^2} + {b^2} + \frac{2}{2}\sqrt {{{\left( {{a^2} + {b^2}} \right)}^2}} \cr & = 2\left( {{a^2} + {b^2}} \right) \cr} $$
∴ Max value - Min value
$$\eqalign{ & = 2\left( {{a^2} + {b^2}} \right) - \left( {a + {b^2}} \right) \cr & = {\left( {a - b} \right)^2} \cr} $$

$$\eqalign{ & {\text{Here, }}z = x\cos \theta + y\sin \theta \cr & {z^2} = {x^2}{\cos ^2}\theta + {y^2}{\sin ^2}\theta + 2\,xy\sin \theta \cos \theta \cr & \Rightarrow 2\,xy\sin \theta \cos \theta = {z^2} - {x^2}{\cos ^2}\theta - {y^2}{\sin ^2}\theta \cr & {\text{Let, }}L = {\left( {x\sin \theta - y\cos \theta } \right)^2} \cr & \Rightarrow L = {x^2}{\sin ^2}\theta + {y^2}{\cos ^2}\theta - 2\,xy\sin \theta \cos \theta \cr & \Rightarrow L = {x^2}{\sin ^2}\theta + {y^2}{\cos ^2}\theta - \left[ {{z^2} - {x^2}{{\cos }^2}\theta - {y^2}{{\sin }^2}\theta } \right] \cr & \Rightarrow L = {x^2}\left[ {{{\sin }^2}\theta + {{\cos }^2}\theta } \right] + {y^2}\left[ {{{\sin }^2}\theta + {{\cos }^2}\theta } \right] - {z^2} \cr & \Rightarrow L = {x^2} + {y^2} - {z^2} \cr} $$

$$\eqalign{ & 3{\left( {\sin \theta - \cos \theta } \right)^4} + 6{\left( {\sin \theta + \cos \theta } \right)^2} + 4\,{\sin ^6}\theta \cr & = 3{\left( {1 - 2\sin \theta \cos \theta } \right)^2} + 6\left( {1 + 2\sin \theta \cos \theta } \right) + 4\,{\sin ^6}\theta \cr & = 3\left( {1 + 4{{\sin }^2}\theta\, {{\cos }^2}\theta - 4\sin \theta \cos \theta } \right) + 6 + 12\sin \theta \cos \theta + 4\,{\sin ^6}\theta \cr & = 9 + 12\,{\sin ^2}\theta\, {\cos ^2}\theta + 4\,{\sin ^6}\theta \cr & = 9 + 12\,{\cos ^2}\theta \left( {1 - {{\cos }^2}\theta } \right) + 4{\left( {1 - {{\cos }^2}\theta } \right)^3} \cr & = 9 + 12\,{\cos ^2}\theta - 12\,{\cos ^4}\theta + 4\left( {1 - {{\cos }^6}\theta - 3\,{{\cos }^2}\theta + 3\,{{\cos }^4}\theta } \right) \cr & = 9 + 4 - 4\,{\cos ^6}\theta \cr & = 13 - 4\,{\cos ^6}\theta \cr} $$