Question

The expression $$\frac{{\cos 6x + 6\cos 4x + 15\cos 2x + 10}}{{\cos 5x + 5\cos 3x + 10\cos x}}$$       is equal to

A. $$\cos 2x$$
B. $$2 \cos x$$  
C. $$\cos^2 x$$
D. $$1 + \cos x$$
Answer :   $$2 \cos x$$
Solution :
The given expression can be written as
$$\eqalign{ & \frac{{\left( {\cos 6x + \cos 4x} \right) + 5\left( {\cos 4x + \cos 2x} \right) + 10\left( {\cos 2x + 1} \right)}}{{\cos 5x + 5\cos 3x + 10\cos x}} \cr & = \frac{{2\cos 5x\cos x + 5 \cdot 2\cos 3x\cos x + 10 \cdot 2{{\cos }^2}x}}{{\cos 5x + 5\cos 3x + 10\cos x}} \cr & = \frac{{2\cos x\left( {\cos 5x + 5\cos 3x + 10\cos x} \right)}}{{\cos 5x + 5\cos 3x + 10\cos x}} = 2\cos x \cr} $$

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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