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

$$\sec A = - \frac{5}{4}.\,{\text{So, }}\frac{\pi }{2} < A < \pi .\,{\text{Hence,}}\tan A = - \frac{3}{4}\,{\text{and cosec}}\,A = \frac{5}{3}.$$
∴ the required equation is $${x^2} - \left( { - \frac{3}{4} + \frac{5}{3}} \right)x + \left( { - \frac{3}{4}} \right) \times \frac{5}{3} = 0.$$

$$\eqalign{ & x = \frac{{3 - \cos 4\theta + 4\sin 2\theta }}{2} \cr & = \frac{{3 - \left( {1 - {{\sin }^2}2\theta } \right) + 4\sin 2\theta }}{2} = {\left( {1 + \sin 2\theta } \right)^2} \cr & y = \frac{{3 - \cos 4\theta - 4\sin 2\theta }}{2} \cr & = \frac{{3 - \left( {1 - {{\sin }^2}2\theta } \right) + 4\sin 2\theta }}{2} = {\left( {1 - \sin 2\theta } \right)^2} \cr & \therefore xy = {\left( {1 - {{\sin }^2}2\theta } \right)^2} = {\cos ^4}2\theta \leqslant 1 \cr} $$

Value $$ = \frac{{2\sin {{18}^ \circ } + 3 - 4{{\cos }^2}{{18}^ \circ }}}{{\cos {{18}^ \circ }}} = \frac{{2\sin {{18}^ \circ } + 3 - 2\left( {1 + \cos {{36}^ \circ }} \right)}}{{\cos {{18}^ \circ }}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2 \cdot \frac{{\sqrt 5 - 1}}{4} + 3 - 2 - 2 \cdot \frac{{\sqrt 5 + 1}}{4}}}{{\cos {{18}^ \circ }}} = 0.$$

$$\eqalign{ & \sec 2x - \tan 2x \cr & = \,\frac{{1 - \sin 2x}}{{\cos 2x}} \cr & = \frac{{1 - \cos 2\left( {\frac{\pi }{4} - x} \right)}}{{\sin 2\left( {\frac{\pi }{4} - x} \right)}} \cr & = \frac{{2{{\sin }^2}\left( {\frac{\pi }{4} - x} \right)}}{{2\sin \left( {\frac{\pi }{4} - x} \right)\cos \left( {\frac{\pi }{4} - x} \right)}} \cr & = \tan \left( {\frac{\pi }{4} - x} \right) \cr} $$

$$\eqalign{ & \cos {20^ \circ } + \cos {100^ \circ } + \cos {140^ \circ } \cr & = \left( {\cos {{140}^ \circ } + \cos {{20}^ \circ }} \right) + \cos {100^ \circ } \cr & = 2\cos \left( {\frac{{{{160}^ \circ }}}{2}} \right) \cdot \cos \left( {\frac{{{{120}^ \circ }}}{2}} \right) + \cos {100^ \circ } \cr & = 2 \cdot \cos {80^ \circ } \cdot \frac{1}{2} + \cos {100^ \circ } \cr & = 2\cos \left( {\frac{{{{180}^ \circ }}}{2}} \right) \cdot \cos \left( {\frac{{{{20}^ \circ }}}{2}} \right) \cr & = 2\cos {90^ \circ } \cdot \cos {10^ \circ } \cr & = 2 \times 0 \times \cos {10^ \circ } = 0 \cr} $$

Here, $$2x = \tan {20^ \circ } + \tan {50^ \circ } = \frac{{\sin {{70}^ \circ }}}{{\cos {{20}^ \circ } \cdot \cos {{50}^ \circ }}} = \sec {50^ \circ } = {\text{cosec}}\,{40^ \circ }$$
and $$2y = \tan {20^ \circ } + \tan {70^ \circ } = \frac{{\sin {{90}^ \circ }}}{{\cos {{20}^ \circ } \cdot \cos {{70}^ \circ }}}$$
$$2y = \frac{1}{{\cos {{20}^ \circ } \cdot \sin {{20}^ \circ }}} = \frac{2}{{\sin {{40}^ \circ }}}.$$

The expression $$ = 2{\sin ^2}\theta - 3\sin 2\theta + 3 = 4 - \left( {\cos 2\theta + 3\sin 2\theta } \right)$$
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4 - \sqrt {{1^2} + {3^2}} \left( {\frac{1}{{\sqrt {10} }}\cos 2\theta + \frac{3}{{\sqrt {10} }}\sin 2\theta } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4 - \sqrt {10} \cos \left( {2\theta - \alpha } \right),\,{\text{where}}\,\,\cos \alpha = \frac{1}{{\sqrt {10} }}. \cr} $$
∴ the least value $$ = 4 - \sqrt {10} \times $$   the greatest value of $$\cos \left( {2\theta - \alpha } \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4 - \sqrt {10} \times 1 = 4 - \sqrt {10} .$$

$$\eqalign{ & \sec \theta - \tan \theta = \lambda \cr & \Rightarrow \,\,\sec \theta + \tan \theta = \frac{1}{\lambda }. \cr} $$
∴ subtracting, $$2\tan \theta = \frac{1}{\lambda } - \lambda \,\,\,\,\,{\text{or, }}2\left( {a - \frac{1}{{4a}}} \right) = \frac{1}{\lambda } - \lambda $$
or, $$2a - \frac{1}{{2a}} = \frac{1}{\lambda } - \lambda $$
$$ \Rightarrow \,\,\lambda = \frac{1}{{2a}}, - 2a.$$

$$\eqalign{ & {\sin ^2}66\frac{{{1^ \circ }}}{2} - {\sin ^2}23\frac{{{1^ \circ }}}{2} \cr & = {\left[ {\sin \left( {{{90}^ \circ } - 23\frac{{{1^ \circ }}}{2}} \right)} \right]^2} - {\sin ^2}23\frac{{{1^ \circ }}}{2} \cr & = {\cos ^2}23\frac{{{1^ \circ }}}{2} - {\sin ^2}23\frac{{{1^ \circ }}}{2} \cr & = \cos 2\left( {23\frac{{{1^ \circ }}}{2}} \right) = \cos {47^ \circ }\,\,\left( {\because \cos 2A = {{\cos }^2}A - {{\sin }^2}A} \right) \cr & = \cos \left[ {2 \times \left( {\frac{{47}}{2}} \right)} \right] = \cos {47^ \circ } \cr} $$

$$\eqalign{ & 3 - 4{\sin ^2}x = y \cr & \therefore \,\,{\sin ^2}x = \frac{{3 - y}}{4}.\,{\text{But }}0 < {\sin ^2}x \leqslant 1\,\,\,\,\,\left( {\because \,\,\sin x = 0\,\,\,\, \Rightarrow \,x = n\pi } \right). \cr & \therefore \,\,0 < \frac{{3 - y}}{4} \leqslant 1\,\,\,{\text{or, }}0 < 3 - y \leqslant 4. \cr} $$