Question

What is $$\frac{{\cos 7x - \cos 3x}}{{\sin 7x - 2\sin 5x + \sin 3x}}$$      equal to ?

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

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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