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

$$\eqalign{ & {\text{Given}},3\,{\cos ^2}A + 2\,{\cos ^2}B = 4 \cr & \Rightarrow 2\,{\cos ^2}B - 1 = 4 - 3\,{\cos ^2}A - 1 \cr & \Rightarrow \cos 2B = 3\left( {1 - {{\cos }^2}A} \right) = 3\,{\sin ^2}A\,\,.....\left( 1 \right) \cr & {\text{and }}2\cos B\sin B = 3\sin A\cos A \cr & \sin 2B = 3\sin A\cos A\,\,.....\left( 2 \right) \cr & {\text{Now}},\cos \left( {A + 2B} \right) = \cos A\cos 2B - \sin A\sin 2B \cr & = \cos A\left( {3\,{{\sin }^2}A} \right) - \sin A\left( {3\sin A\cos A} \right) = 0\left[ {{\text{using eqs}}{\text{. }}\left( 1 \right){\text{and}}\left( 2 \right)} \right] \cr & \Rightarrow A + 2B = \frac{\pi }{2} \cr} $$

$$\eqalign{ & \cos \alpha + \cos \beta = - a,\cos \alpha \cdot \cos \beta = b\,\,{\text{and }}\sin \alpha + \sin \beta = - p,\sin \alpha \cdot \sin \beta = q. \cr & \therefore \,\,{a^2} + {p^2} = 2 + 2\cos \left( {\alpha - \beta } \right)\,\,{\text{and }}b + q = \cos \left( {\alpha - \beta } \right). \cr & \therefore \,\,{a^2} + {p^2} = 2 + 2\left( {b + q} \right). \cr} $$

The given equation is
$$\left( {\cos p - 1} \right){x^2} + \left( {\cos p} \right)x + \sin p = 0$$
For this equation to have real roots $$D \geqslant 0$$
$$\eqalign{ & \Rightarrow \,\,{\cos ^2}p - 4\sin p\left( {\cos p - 1} \right) \geqslant 0 \cr & \Rightarrow \,\,{\cos ^2}p - 4\sin p\cos p + 4{\sin ^2}p + 4\sin p - 4{\sin ^2}p \geqslant 0 \cr & \Rightarrow \,\,{\left( {\cos p - 2\sin p} \right)^2} + 4\sin p\left( {1 - \sin p} \right) \geqslant 0 \cr} $$
For every real value of $$p{\left( {\cos p - 2\sin p} \right)^2} \geqslant 0$$     and
$$1 - \sin p \geqslant 0\,\,\,\,\,\,\,\,\therefore \,\,D \geqslant 0,\,\,\forall \,p \in \left( {0,\pi } \right)$$

If $$\left( {1 + \sin \alpha } \right)\left( {1 + \sin \beta } \right)\left( {1 + \sin \gamma } \right) = k$$
And $$\left( {1 - \sin \alpha } \right)\left( {1 - \sin \beta } \right)\left( {1 - \sin \gamma } \right) = k$$
The value of $${k^2} = k \cdot k.$$
$$\eqalign{ & = \left( {1 + \sin \alpha } \right)\left( {1 + \sin \beta } \right)\left( {1 + \sin \gamma } \right)\left( {1 - \sin \alpha } \right)\left( {1 - \sin \beta } \right)\left( {1 - \sin \gamma } \right) \cr & = \left( {1 + \sin \alpha } \right)\left( {1 - \sin \alpha } \right)\left( {1 + \sin \beta } \right)\left( {1 - \sin \beta } \right)\left( {1 + \sin \gamma } \right)\left( {1 - \sin \gamma } \right) \cr & = \left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\sin }^2}\beta } \right)\left( {1 - {{\sin }^2}\gamma } \right) \cr & \Rightarrow {k^2} = {\cos ^2} \alpha \,{\cos ^2} \beta \,{\cos ^2} \gamma \cr & \therefore k = + \cos \alpha \cos \beta \cos \gamma . \cr} $$

$$\eqalign{ & L = \frac{{1 - \tan {2^ \circ }\cot {{62}^ \circ }}}{{\tan {{152}^ \circ } - \cot {{88}^ \circ }}} = \frac{{1 - \tan {2^ \circ }\cot {{\left( {90 - 28} \right)}^ \circ }}}{{\tan {{\left( {180 - 28} \right)}^ \circ } - \cot {{\left( {90 - 2} \right)}^ \circ }}} \cr & \Rightarrow L = \frac{{1 - \tan {2^ \circ }\tan {{28}^ \circ }}}{{ - \tan {{28}^ \circ } - \tan {2^ \circ }}} = - \left[ {\frac{{1 - \tan {2^ \circ }\tan {{28}^ \circ }}}{{\tan {2^ \circ } + \tan {{28}^ \circ }}}} \right] \cr & \Rightarrow L = - \frac{1}{{\tan {{\left( {2 + 28} \right)}^ \circ }}} = - \frac{1}{{\tan {{30}^ \circ }}} = - \sqrt 3 \,\,\,\,\left[ {\because \tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right] \cr} $$

$$\eqalign{ & \sin A \bullet \sin \left( {{{60}^ \circ } - A} \right)\sin \left( {{{60}^ \circ } + A} \right) = k\sin 3A \cr & \Rightarrow \sin A \cdot \frac{{\sin 3A}}{{4\sin A}} = k \cdot \sin 3A\,\,\,\left[ {\because \sin \left( {{{60}^ \circ } + A} \right) \cdot \sin \left( {{{60}^ \circ } - A} \right) = \frac{{\sin 3A}}{{4\sin A}}} \right] \cr & \Rightarrow \frac{{\sin 3A}}{4} = k \cdot \sin 3A \cr & \therefore k = \frac{1}{4} \cr} $$

To simplify the det. Let $$\sin x = a;\cos x = b$$     the equation becomes
\[\begin{array}{l} \left| {\begin{array}{*{20}{c}} a&b&b\\ b&a&b\\ b&b&a \end{array}} \right| = 0\,\,{\rm{Operating }}\,\,{C_2} - {C_1};{C_3} - {C_2}\,{\rm{we\, get}}\\ \left| {\begin{array}{*{20}{c}} a&{b - a}&0\\ b&{a - b}&{b - a}\\ b&0&{a - b} \end{array}} \right| = 0 \end{array}\]
$$\eqalign{ & \Rightarrow \,\,a{\left( {a - b} \right)^2} - \left( {b - a} \right)\left[ {b\left( {a - b} \right) - b\left( {b - a} \right)} \right] = 0 \cr & \Rightarrow \,\,a{\left( {a - b} \right)^2} - 2b\left( {b - a} \right)\left( {a - b} \right) = 0 \cr & \Rightarrow \,\,{\left( {a - b} \right)^2}\left( {a - 2b} \right) = 0 \cr & \Rightarrow \,\,\left( {a = b} \right){\text{or }}a = 2b \cr & \Rightarrow \,\,\frac{a}{b} = 1\,{\text{or }}\frac{a}{b} = 2 \cr & \Rightarrow \,\,\tan x = 1\,\,{\text{or }}\tan x = 2.\,{\text{But we have }} - \frac{\pi }{4} \leqslant x \leqslant \frac{\pi }{4} \cr & \Rightarrow \,\,\tan \left( { - \frac{\pi }{4}} \right) \leqslant \tan x \leqslant \tan \left( {\frac{\pi }{4}} \right) \cr & \Rightarrow \,\, - 1 \leqslant \tan x \leqslant 1 \cr & \therefore \,\,\tan x = 1 \cr & \Rightarrow \,\,x = \frac{\pi }{4} \cr} $$
∴ Only one real root is there.

$$\eqalign{ & {\text{Here, }}{\cos ^2}x + \cos x - 1 = 0\,\,\,{\text{or, }}\cos x = \frac{{ - 1 + \sqrt 5 }}{2}\left\{ {\because \,\,\cos x \ne \frac{{ - 1 - \sqrt 5 }}{2} < - 1} \right\} \cr & \therefore \,\,{\cos ^2}x = {\left( {\frac{{\sqrt 5 - 1}}{2}} \right)^2} = \frac{{6 - 2\sqrt 5 }}{4} = \frac{{3 - \sqrt 5 }}{2} \cr & \therefore \,\,{\text{value}} = \left( {1 - \frac{{ 3 - \sqrt 5}}{2}} \right)\left( {2 - \frac{{3 - \sqrt 5 }}{2}} \right) = \frac{{\sqrt 5 - 1}}{2} \cdot \frac{{\sqrt 5 + 1}}{2} = 1. \cr} $$

$${\left( {x + y} \right)^2} - 4xy = {\left( {x - y} \right)^2} \geqslant 0$$
$$\therefore \,\,{\left( {x + y} \right)^2} \geqslant 4xy$$    and equality holds when $$x = y.$$
But $${\sec ^2}\theta \geqslant 1.$$   Hence, $${\sec ^2}\theta = \frac{{4xy}}{{{{\left( {x + y} \right)}^2}}}$$    can hold only if $${\left( {x + y} \right)^2} = 4xy.$$

$$\eqalign{ & \because \,\,\tan 3\theta - \tan 2\theta - \tan \theta = \tan 3\theta \cdot \tan 2\theta \cdot \tan \theta \cr & \therefore \,\,0 = ab\tan 3\theta \cr & \Rightarrow \,\,\tan 3\theta = 0 \cr & \Rightarrow \,\,\tan \theta + \tan 2\theta = 0 \cr & \Rightarrow \,\,a + b = 0. \cr} $$