Question

What is $$\cos {20^ \circ } + \cos {100^ \circ } + \cos {140^ \circ }$$      equal to ?

A. $$2$$
B. $$1$$
C. $$\frac{1}{2}$$
D. $$0$$  
Answer :   $$0$$
Solution :
$$\eqalign{ & \cos {20^ \circ } + \cos {100^ \circ } + \cos {140^ \circ } \cr & = \left( {\cos {{140}^ \circ } + \cos {{20}^ \circ }} \right) + \cos {100^ \circ } \cr & = 2\cos \left( {\frac{{{{160}^ \circ }}}{2}} \right) \cdot \cos \left( {\frac{{{{120}^ \circ }}}{2}} \right) + \cos {100^ \circ } \cr & = 2 \cdot \cos {80^ \circ } \cdot \frac{1}{2} + \cos {100^ \circ } \cr & = 2\cos \left( {\frac{{{{180}^ \circ }}}{2}} \right) \cdot \cos \left( {\frac{{{{20}^ \circ }}}{2}} \right) \cr & = 2\cos {90^ \circ } \cdot \cos {10^ \circ } \cr & = 2 \times 0 \times \cos {10^ \circ } = 0 \cr} $$

Releted MCQ Question on
Trigonometry >> Trigonometric Ratio and Identities

Releted Question 1

If $$\tan \theta = - \frac{4}{3},$$   then $$\sin \theta $$  is

A. $$ - \frac{4}{5}{\text{ but not }}\frac{4}{5}$$
B. $$ - \frac{4}{5}{\text{ or }}\frac{4}{5}$$
C. $$ \frac{4}{5}{\text{ but not }} - \frac{4}{5}$$
D. None of these
Releted Question 2

If $$\alpha + \beta + \gamma = 2\pi ,$$    then

A. $$\tan \frac{\alpha }{2} + \tan \frac{ \beta }{2} + \tan \frac{\gamma }{2} = \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$$
B. $$\tan \frac{\alpha }{2}\tan \frac{\beta }{2} + \tan \frac{\beta }{2}\tan \frac{\gamma }{2} + \tan \frac{\gamma }{2}\tan \frac{\alpha }{2} = 1$$
C. $$\tan \frac{\alpha }{2} + \tan \frac{ \beta }{2} + \tan \frac{\gamma }{2} = - \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$$
D. None of these
Releted Question 3

Given $$A = {\sin ^2}\theta + {\cos ^4}\theta $$    then for all real values of $$\theta $$

A. $$1 \leqslant A \leqslant 2$$
B. $$\frac{3}{4} \leqslant A \leqslant 1$$
C. $$\frac{13}{16} \leqslant A \leqslant 1$$
D. $$\frac{3}{4} \leqslant A \leqslant \frac{{13}}{{16}}$$
Releted Question 4

The value of the expression $$\sqrt 3 \,{\text{cosec}}\,{\text{2}}{{\text{0}}^ \circ } - \sec {20^ \circ }$$     is equal to

A. 2
B. $$\frac{{2\sin {{20}^ \circ }}}{{\sin {{40}^ \circ }}}$$
C. 4
D. $$\frac{{4\sin {{20}^ \circ }}}{{\sin {{40}^ \circ }}}$$

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