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

$$\eqalign{ & \sin x + \sin y = a \cr & \Rightarrow 2\sin \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right) = a\,\,\,.....\left( 1 \right) \cr & \cos x + \cos y = b \cr & \Rightarrow 2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right) = b\,\,\,.....\left( 2 \right) \cr} $$
dividing eq $$\left( 1 \right)\,\,\& \,\,\left( 2 \right)$$
$$\tan \left( {\frac{{x + y}}{2}} \right) = \frac{a}{b}$$
Squaring of eq (1) $$\&$$ (2) and adding -
$$\eqalign{ & 4\,{\cos ^2}\left( {\frac{{x - y}}{2}} \right) = {a^2} + {b^2} \cr & {\sec ^2}\left( {\frac{{x - y}}{2}} \right) = \frac{4}{{{a^2} + {b^2}}}\, \cr & {\text{again}} , \cr & {\tan ^2}\left( {\frac{{x + y}}{2}} \right) + {\tan ^2}\left( {\frac{{x - y}}{2}} \right) \cr & = {\left( {\frac{a}{b}} \right)^2} + {\sec ^2}\left( {\frac{{x - y}}{2}} \right) - 1 \cr & = \frac{{{a^2}}}{{{b^2}}} + \frac{4}{{{a^2} + {b^2}}} - 1 \cr & = \frac{{{a^4} - {b^4} + 4{b^2}}}{{{a^2}{b^2} + {b^4}}} \cr} $$

$$\eqalign{ & {\left( {3\sin \theta + 4\cos \theta } \right)^2} + {\left( {4\sin \theta - 3\cos \theta } \right)^2} = 16 + 9 = 25 \cr & \therefore \,\,25 + {\left( {4\sin \theta - 3\cos \theta } \right)^2} = 25 \cr & \therefore \,\,4\sin \theta - 3\cos \theta = 0. \cr} $$

$$\eqalign{ & A = {\sin ^2}x + {\cos ^4}x \cr & = {\sin ^2}x + {\cos ^2}x\left( {1 - {{\sin }^2}x} \right) \cr & = {\sin ^2}x + {\cos ^2}x - \frac{1}{4}{\left( {2\sin x.\cos x} \right)^2} \cr & = 1 - \frac{1}{4}{\sin ^2}\left( {2x} \right) \cr & {\text{Now }}0 \leqslant {\sin ^2}\left( {2x} \right) \leqslant 1 \cr & \Rightarrow \,\,0 \geqslant - \frac{1}{4}{\sin ^2}\left( {2x} \right) \geqslant - \frac{1}{4} \cr & \Rightarrow \,\,1 \geqslant 1 - \frac{1}{4}{\sin ^2}\left( {2x} \right) \geqslant 1 - \frac{1}{4} \cr & \Rightarrow \,\,1 \geqslant A \geqslant \frac{3}{4} \cr} $$

$$\eqalign{ & A = {\sin ^2}\theta + {\cos ^4}\theta \cr & = {\sin ^2}\theta + {\left( {1 - {{\sin }^2}\theta } \right)^2} \cr & = {\sin ^4}\theta - {\sin ^2}\theta + 1 \cr & \Rightarrow \,\,A = {\left( {{{\sin }^2}\theta - \frac{1}{2}} \right)^2} + \frac{3}{4} \cr & {\text{But }}0 \leqslant {\left( {{{\sin }^2}\theta - \frac{1}{2}} \right)^2} \leqslant \frac{1}{4} \cr & \therefore \,\,\frac{3}{4} \leqslant {\left( {{{\sin }^2}\theta - \frac{1}{2}} \right)^2} + \frac{3}{4} \leqslant 1\,\,{\text{or }}\frac{3}{4} \leqslant A \leqslant 1 \cr} $$

From Sine Rule

$$\eqalign{ & \frac{{{\text{AB}}}}{{\sin \theta }} = \frac{{\sqrt {{p^2} + {q^2}} }}{{\sin \left( {\pi - \left( {\theta + \alpha } \right)} \right)}} \cr & {\text{AB}} = \frac{{\sqrt {{p^2} + {q^2}} \sin \theta }}{{\sin \theta \cos \alpha + \cos \theta \sin \alpha }} \cr & = \frac{{\left( {{p^2} + {q^2}} \right)\sin \theta }}{{q\sin \theta + p\cos \theta }} \cr & \left( {\because \,\,\cos \alpha = \frac{q}{{\sqrt {{p^2} + {q^2}} }}\,{\text{and sin}}\alpha = \frac{p}{{\sqrt {{p^2} + {q^2}} }}} \right) \cr} $$

We have $${\sec ^2}\theta = \frac{{4xy}}{{{{\left( {x + y} \right)}^2}}}$$
But $${\sec ^2}\theta \geqslant 1$$
$$\eqalign{ & \Rightarrow \,\,\frac{{4xy}}{{{{\left( {x + y} \right)}^2}}} \geqslant 1 \cr & \Rightarrow \,\,4xy \geqslant {x^2} + {y^2} + 2xy \cr & \Rightarrow \,\,{x^2} + {y^2} - 2xy \leqslant 0 \cr & \Rightarrow \,\,{\left( {x - y} \right)^2} \leqslant 0 \cr} $$
⇒ $$x - y = 0$$   [as perfect square of real number can never be negative]
Also then $$x \ne 0$$  as then $${\sec ^2}\theta $$  will become indeterminate.

$$\left[ {1 + \cos \frac{\pi }{8}} \right]\left[ {1 + \cos \frac{{3\pi }}{8}} \right]\left[ {1 + \cos \frac{{5\pi }}{8}} \right]\left[ {1 + \frac{{\cos 7\pi }}{8}} \right]$$
We have,
$$\eqalign{ & \cos \frac{{7\pi }}{8} = \cos \left[ {\pi - \frac{\pi }{8}} \right] = - \cos \frac{\pi }{8}\,{\text{and}}\,\,{{\cos}}\frac{{5\pi }}{8} = \cos \left[ {\pi - \frac{{3\pi }}{8}} \right] = - \cos \frac{{3\pi }}{8} \cr & \therefore \left[ {1 + \cos \frac{\pi }{8}} \right]\left[ {1 + \cos \frac{{3\pi }}{8}} \right]\left[ {1 - \cos \frac{\pi }{8}} \right]\left[ {1 - \cos \frac{{3\pi }}{8}} \right] \cr & = \left[ {1 - {{\cos }^2}\frac{\pi }{8}} \right]\left[ {1 - {{\cos }^2}\frac{{3\pi }}{8}} \right] = {\sin ^2}\frac{\pi }{8} \cdot {\sin ^2}\frac{{3\pi }}{8} \cr & = \frac{1}{4}\left[ {2\,{{\sin }^2}\frac{\pi }{8} \cdot 2\,{{\sin }^2}\frac{{3\pi }}{8}} \right] \cr & = \frac{1}{4}\left[ {\left( {1 - \cos \frac{\pi }{4}} \right)\left( {1 - \cos \frac{{3\pi }}{4}} \right)} \right]\,\,\,\,\,\,\,\,\,\left( {\because 1 - \cos \theta = 2\,{{\sin }^2}\frac{\theta }{2}} \right) \cr & = \frac{1}{4}\left[ {\left( {1 - \frac{1}{{\sqrt 2 }}} \right)\left( {1 + \frac{1}{{\sqrt 2 }}} \right)} \right] = \frac{1}{8} \cr} $$

Since, $$A$$ and $$B$$ are complementary angles, then $$A + B = {90^ \circ }$$
Now, $$\cos A\cos B = \cos A\cos \left( {{{90}^ \circ } - A} \right)$$
$$ = \cos A\sin A = \frac{1}{2}\sin 2A$$
Since, $$ - 1 \leqslant \sin 2A \leqslant 1$$
Hence, $$ - \frac{1}{2} \leqslant \frac{1}{2}\sin 2A \leqslant \frac{1}{2}.$$
Thus, greatest and least values of $$\cos A \cos B$$   are $$ \frac{1}{2}$$ and $$ - \frac{1}{2} .$$

$$\eqalign{ & \frac{{\cos 7x - \cos 3x}}{{\sin 7x - 2\sin 5x + \sin 3x}} \cr & = \frac{{ - 2\sin \frac{{7x + 3x}}{2} \cdot \sin \frac{{7x - 3x}}{2}}}{{2\sin \frac{{7x + 3x}}{2} \cdot \cos \frac{{7x - 3x}}{2} - 2\sin 5x}} \cr} $$
\[\left( \begin{gathered} \because \sin C + \sin D = 2\sin \left( {\frac{{C + D}}{2}} \right) \cdot \cos \left( {\frac{{C - D}}{2}} \right) \hfill \\ {\text{and}}\,\,\cos C - \cos D = - 2\sin \left( {\frac{{C + D}}{2}} \right)\sin \left( {\frac{{C - D}}{2}} \right) \hfill \\ \end{gathered} \right)\]
$$\eqalign{ & = \frac{{ - 2\sin 5x \cdot \sin 2x}}{{2\sin 5x\cos 2x - 2\sin 5x}} \cr & = \frac{{ - 2\sin 5x \cdot \sin 2x}}{{ - 2\sin 5x\left[ {1 - \cos 2x} \right]}} \cr & = \frac{{\sin 2x}}{{1 - 1 + 2\,{{\sin }^2}x}}\,\,\,\left( {\because \cos 2x = 1 - 2\,{{\sin }^2}x} \right) \cr & = \frac{{2\sin x\cos x}}{{2\,{{\sin }^2}x}} = \cot x \cr} $$

$$\frac{{\sin \left( {x + y} \right)}}{{\sin \left( {x - y} \right)}} = \frac{{a + b}}{{a - b}}$$
Applying componendo and dividendo, we get
$$\eqalign{ & \frac{{\sin \left( {x + y} \right) + \sin \left( {x - y} \right)}}{{\sin \left( {x + y} \right) - \sin \left( {x - y} \right)}} = \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{{\left( {a + b} \right) - \left( {a - b} \right)}} \cr & \Rightarrow \frac{{2\sin x \cdot \cos y}}{{2\cos x \cdot \cos y}} = \frac{{2a}}{{2b}} \cr & \Rightarrow \tan x \cdot \cot y = \frac{a}{b} \cr & \therefore \frac{{\tan x}}{{\tan y}} = \frac{a}{b} \cr} $$