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

$$\eqalign{ & \frac{{{{\sin }^3}A + \sin 3A}}{{\sin A}} + \frac{{{{\cos }^3}A - \cos 3A}}{{\cos A}} \cr & \Rightarrow \frac{{{{\sin }^3}A + 3\,\sin A - 4\,{{\sin }^3}A}}{{\sin A}} + \frac{{{{\cos }^3}A - \left[ {4\,{{\cos }^3}A - 3\cos A} \right]}}{{\cos A}} \cr & \Rightarrow \frac{{3\sin A - 3\,{{\sin }^3}A}}{{\sin A}} + \frac{{\left( { - 3\,{{\cos }^3}A + 3\cos A} \right)}}{{\cos A}} \cr & = 3 - 3\,{\sin ^2}A - 3\,{\cos ^2}A + 3 \cr & = 6 - 3\left( {{{\cos }^2}A + {{\sin }^2}A} \right) \cr & = 6 - 3\left( 1 \right) \cr & = 3 \cr} $$

Given, $$\cos x + \sin x = \frac{1}{2}$$
$$ \Rightarrow 1 + \sin 2x = \frac{1}{4}$$
$$ \Rightarrow \sin 2x = - \frac{3}{4},$$    so $$x$$ is obtuse and $$\frac{{2\tan x}}{{1 + {{\tan }^2}x}} = - \frac{3}{4}$$
$$\eqalign{ & \Rightarrow 3\tan ^2 x + 8\tan x + 3 = 0 \cr & \therefore \tan x = \frac{{ - 8 \pm \sqrt {64 - 36} }}{6} = - \frac{{ - 4 \pm \sqrt 7 }}{3} \cr & {\text{as }}\tan x < 0 \cr & \therefore \tan x = \frac{{ - 4 - \sqrt 7 }}{3} \cr} $$

$$\eqalign{ & 2{\sin ^2}\theta - 5\sin \theta + 2 > 0 \cr & \Rightarrow \,\,\left( {\sin \theta - 2} \right)\left( {2\sin \theta - 1} \right) > 0 \cr & \Rightarrow \,\,\sin \theta < \frac{1}{2}\,\,\,\,\,\,\left[ {\because \,\, - 1 \leqslant \sin \theta \leqslant 1} \right] \cr} $$

From graph, we get $$x \in \left( {0,\frac{\pi }{6}} \right) \cup \left( {\frac{{5\pi }}{6},2\pi } \right)$$

$$\eqalign{ & \because \,\,\theta \in \left( {0,\frac{\pi }{4}} \right) \cr & \Rightarrow \,\,\tan \theta < 1\,{\text{and }}\cot \theta > 1 \cr & {\text{Let tan}}\theta = 1 - x\,{\text{and cot}}\theta = 1 + y \cr} $$
Where $$x, y > 0$$   and are very small, then
$$\eqalign{ & \therefore \,\,{t_1} = {\left( {1 - x} \right)^{1 - x}},{t_2} = {\left( {1 - x} \right)^{1 + y}}, \cr & {t_3} = {\left( {1 + y} \right)^{1 - x}},{t_4} = {\left( {1 + y} \right)^{1 + y}} \cr} $$
$${\text{Clearly, }}{t_4} > {t_3}\,{\text{and }}{t_1} > {t_2}\,{\text{also, }}{t_3} > {t_1}$$         NOTE THIS STEP
$${\text{Thus}}\,{\text{ }}{t_4} > {t_3} > {t_1} > {t_2}.$$

Use $$\cos \left( {\alpha + \beta } \right) = \frac{{1 - {{\tan }^2}\frac{{\alpha + \beta }}{2}}}{{1 + {{\tan }^2}\frac{{\alpha + \beta }}{2}}},\tan \frac{{\alpha + \beta }}{2} = \frac{{\tan \frac{\alpha }{2} + \tan \frac{\beta }{2}}}{{1 - \tan \frac{\alpha }{2} \cdot \tan \frac{\beta }{2}}}$$
and $$\tan \frac{\alpha }{2} + \tan \frac{\beta }{2} = \frac{{13}}{4},\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = \frac{{15}}{8}.$$

$$\eqalign{ & \frac{1}{{\sin x}} - 1 \geqslant 0;\,\,\frac{{1 - \sin x}}{{\sin x}} \geqslant 0 \cr & \frac{{\sin x - 1}}{{\sin x}} \leqslant 0 \cr & \Rightarrow 0 < \sin x \leqslant 1 \cr & \Rightarrow x \in \mathop \cup \limits_{n \in I} \left[ {2n\pi ,\left( {2n + 1} \right)\pi } \right] \cr} $$

$$\eqalign{ & {\sin ^2}{5^ \circ } + {\sin ^2}{10^ \circ } + {\sin ^2}{15^ \circ } + ..... + ..... + {\sin ^2}{75^ \circ } + {\sin ^2}{80^ \circ } + {\sin ^2}{85^ \circ } + {\sin ^2}{90^ \circ } \cr & \Rightarrow {\sin ^2}{5^ \circ } + {\sin ^2}{10^ \circ } + {\sin ^2}{15^ \circ } + ..... +{\sin ^2}\left( {{{90}^ \circ} - {{15}^ \circ }} \right) + {\sin ^2}\left( {{{90}^ \circ} - {{10}^ \circ}} \right) + {\sin ^2}\left( {{{90}^ \circ} - {{5}^ \circ}} \right) + 1 \cr & \Rightarrow {\sin ^2}{{5}^ \circ} + {\sin ^2}{{10}^ \circ} + {\sin ^2}{15^ \circ } + ..... + {\cos ^2}{15^ \circ } + {\cos ^2}{10^ \circ } + {\cos ^2}{5^ \circ } + 1 \cr & \Rightarrow \left( {1 + 1 + 1 + .....8\,{\text{times}}} \right) + \sin {45^ \circ } + 1 \cr & \Rightarrow 8 + \frac{1}{2} + 1 = \frac{{19}}{2} \cr} $$

$$\eqalign{ & \sec \alpha + {\text{cosec}}\,\alpha = p\,\,{\text{and}}\sec \alpha \cdot {\text{cosec}}\,\alpha = q. \cr & \therefore \,\,\sin\alpha + \cos\alpha = p\sin \alpha \cdot \cos \alpha \,\,{\text{and }}\sin \alpha \cdot \cos \alpha = \frac{1}{q}. \cr & \therefore \,\, \sin \alpha + \cos \alpha = \frac{p}{q} \cr & \therefore \,\,\frac{{{p^2}}}{{{q^2}}} = 1 + 2\sin \alpha \cdot \cos \alpha = 1 + \frac{2}{q}. \cr} $$

$$\eqalign{ & {\text{We have, }}\sin \theta = \frac{{12}}{{13}} \cr & \cos \theta = \sqrt {1 - {{\sin }^2}\theta } = \sqrt {1 - {{\left( {\frac{{12}}{{13}}} \right)}^2}} = \frac{5}{{13}} \cr & {\text{and }}\cos \phi = \frac{{ - 3}}{5}, \sin\phi = \sqrt {1 - \frac{9}{{25}}} = \frac{{ - 4}}{5},\left[ {\because \pi < \phi < \frac{{3\pi }}{2}} \right] \cr & {\text{Now, }}\sin \left( {\theta + \phi } \right) = \sin \theta \cdot \cos \phi + \cos \theta \cdot \sin \phi \cr & = \left( {\frac{{12}}{{13}}} \right)\left( {\frac{{ - 3}}{5}} \right) + \left( {\frac{5}{{13}}} \right)\left( {\frac{{ - 4}}{5}} \right) = \frac{{ - 36}}{{65}} - \frac{{20}}{{65}} = \frac{{ - 56}}{{65}} \cr} $$

$$\eqalign{ & f\left( \theta \right) = \sin \theta \left( {\sin \theta + \sin 3\theta } \right) \cr & = \left( {\sin \theta + 3\sin \theta - 4{{\sin }^3}\theta } \right).\sin \theta \cr & = \left( {4\sin \theta - 4{{\sin }^3}\theta } \right)\sin \theta = {\sin ^2}\theta \left( {4 - 4{{\sin }^2}\theta } \right) \cr & = 4{\sin ^2}\theta \left( {1 - {{\sin }^2}\theta } \right) \cr & = 4{\sin ^2}\theta {\cos ^2}\theta \cr & = {\left( {2\sin \theta \cos \theta } \right)^2} \cr & = {\left( {\sin 2\theta } \right)^2} \geqslant 0 \cr} $$
which is true for all $$\theta .$$