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

We have two functions $$f\left( x \right)$$  and $$g\left( x \right)$$  have periods $$T_1$$ and $$T_2$$ respectively, then each of $$f\left( x \right) \pm g\left( x \right);f\left( x \right) \cdot g\left( x \right);\frac{{f\left( x \right)}}{{g\left( x \right)}},$$       provided $$g\left( x \right) = 0$$   has period equal to the LCM of $$T_1$$ and $$T_2 .$$
Now, we know that $$\sin x$$  or $$\cos x$$  has period $$2\pi .$$
Hence, period of $$y = \sin t + \cos 2t{\text{ is }}2\pi .$$

Use $$\sin \alpha + sin\left( {\alpha + \beta } \right) + \sin \left( {\alpha + 2\beta } \right) + ..... + \sin \left( {\alpha + \overline {n - 1} \beta } \right)$$
$$\eqalign{ & \,\,\, = \frac{{\sin \frac{{n\beta }}{2}}}{{\sin \frac{\beta }{2}}}\sin \frac{{2\alpha + \left( {n - 1} \right)\beta }}{2}. \cr & {\text{Here, }}\alpha = \frac{\pi }{n},\beta = \frac{{2\pi }}{n}. \cr} $$

$$\eqalign{ & \cot \,\theta - 2\,\cot \,2\theta \cr & = \frac{{\cos \,\theta }}{{\sin \,\theta }} - \frac{{2\,\cos \,2\theta }}{{\sin \,2\theta }} \cr & = \frac{{\sin \,\theta .\cos \,\theta - 2\,\cos \,2\theta .\sin \,\theta }}{{\sin \,\theta .\sin \,2\theta }} \cr & = \frac{{\sin \left( {2\theta - \theta } \right) - \cos \,2\theta .\sin \,\theta }}{{\sin \,\theta .\sin \,2\theta }} \cr & = \frac{{\sin \,\theta \left\{ {1 - 1 + 2\,{{\sin }^2}\theta } \right\}}}{{\sin \,\theta .2\,\sin \,\theta .\cos \,\theta }} \cr & = \tan \,\theta \cr & {\text{Hence, option B is the correct option}}{\text{.}} \cr} $$

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

$$\eqalign{ & {p_n} - {p_{n - 2}} = \left( {{{\cos }^n}\theta + {{\sin }^n}\theta } \right) - \left( {{{\cos }^{n - 2}}\theta + {{\sin }^{n - 2}}\theta } \right) \cr & = {\cos ^{n - 2}}\theta \left( {{{\cos }^2}\theta - 1} \right) + {\sin ^{n - 2}}\theta \left( {{{\sin }^2}\theta - 1} \right) \cr & = - {\sin ^2}\theta \,{\cos ^{n - 2}}\theta - {\cos ^2}\theta \,{\sin ^{n - 2}}\theta \cr & = - {\sin ^2}\theta \,{\cos ^2}\theta \left( {{{\cos }^{n - 4}}\theta + {{\sin }^{n - 4}}\theta } \right) \cr & = - {\sin ^2}\theta \,{\cos ^2}\theta {p_{n - 4}} = k{p_{n - 4}} \cr & \Rightarrow k = - {\sin ^2}\theta \,{\cos ^2}\theta \cr} $$

$$\eqalign{ & {\sin ^2}\left( {3\pi } \right) + {\cos ^2}\left( {4\pi } \right) + {\tan ^2}\left( {5\pi } \right) \cr & = {\sin ^2}\left( {3\pi } \right) + {\cos ^2}\left( {\pi + 3\pi } \right) + {\tan ^2}\left( {5\pi } \right) \cr & = \left( {{{\sin }^2}\left( {3\pi } \right) + {{\cos }^2}\left( {3\pi } \right)} \right) + {\tan ^2}\left( {2 \times 2\pi + \pi } \right) \cr & = 1 + {\tan ^2}\pi = {\sec ^2}\pi = 1 \cr} $$

$$\sqrt {2 + \sqrt {2 + \sqrt {2 + ..... + \sqrt {2\left( {1 + \cos \theta } \right)} } } } ,$$        there being $$n$$ number of $$2’s$$
$$ = \sqrt {2 + \sqrt {2 + \sqrt {2 + ..... + \sqrt {2\left( {1 + \cos \frac{\theta }{2}} \right)} } } } ,$$         there being $$(n - 1)$$  number of $$2’s$$
$$ = \sqrt {2 + \sqrt {2 + \sqrt {2 + ..... + \sqrt {2\left( {1 + \cos \frac{\theta }{{{2^2}}}} \right)} } } } ,$$         there being $$(n - 2)$$  number of $$2’s$$
$$ = ..... = \sqrt {2\left( {1 + \cos \frac{\theta }{{{2^{n - 1}}}}} \right)} = 2\cos \frac{\theta }{{{2^n}}}.$$

$$\eqalign{ & 2{\cos ^2}\theta + \cos \theta + \left( {k - 1} \right) = 0 \cr & \therefore \,\,\cos \theta = \frac{{ - 1 \pm \sqrt {1 - 8\left( {k - 1} \right)} }}{4} = \frac{{ - 1 \pm \sqrt {9 - 8k} }}{4}. \cr} $$
For real, $$\cos\theta ,9 - 8k \geqslant 0,\,{\text{i}}{\text{.e}}{\text{., }}k \leqslant \frac{9}{8}.$$
Also, $$ - 1 \leqslant \frac{{ - 1 \pm \sqrt {9 - 8k} }}{4} \leqslant 1\,\,{\text{or, }} - 4 \leqslant - 1 \pm \sqrt {9 - 8k} \leqslant 4$$
or, $$ - 3 \leqslant \pm \sqrt {9 - 8k} \leqslant 5$$
$$\eqalign{ & \therefore \,\, - 3 \leqslant - \sqrt {9 - 8k} \,\,{\text{and }}\sqrt {9 - 8k} \leqslant 5 \cr & \Rightarrow \,\,9 - 8k \leqslant 9\,\,{\text{and }}9 - 8k \leqslant 25 \cr & \therefore \,\,k \geqslant 0\,\,{\text{and }}k \geqslant - 2 \cr & \Rightarrow \,\,k \geqslant 0. \cr} $$

$$\tan \frac{\alpha }{2} + \cot \frac{\alpha }{2} = \frac{1}{{\cos \frac{\alpha }{2} \cdot \sin \frac{\alpha }{2}}} = \frac{2}{{\sin \alpha }} = \frac{2}{p}\,{\text{and}}\tan \frac{\alpha }{2} \cdot \cot \frac{\alpha }{2} = 1.$$
∴ the required equation is $${x^2} - \frac{2}{p}x + 1 = 0.$$

For all $$x,\cos x \geqslant - 1$$
$$\therefore \cos A + \cos B + \cos C \geqslant - 3,$$      and equality holds if
$$\cos A = \cos B = \cos C = - 1,$$      which can be attained if $$A = \pi ,B = \pi ,C = - \pi $$