Question

The value of $$\sum\limits_{k = 1}^{13} {\frac{1}{{\sin \left( {\frac{\pi }{4} + \frac{{\left( {k - 1} \right)\pi }}{6}} \right)\sin \left( {\frac{\pi }{4} + \frac{{k\pi }}{6}} \right)}}} $$       is equal to

A. $$3 - \sqrt 3 $$
B. $$2\left( {3 - \sqrt 3 } \right)$$
C. $$2\left( {\sqrt 3 - 1} \right)$$  
D. $$2\left( {2 - \sqrt 3 } \right)$$
Answer :   $$2\left( {\sqrt 3 - 1} \right)$$
Solution :
$$\eqalign{ & \sum\limits_{k = 1}^{13} {\frac{1}{{\sin \left( {\frac{\pi }{4} + \frac{{\left( {k - 1} \right)\pi }}{6}} \right)\sin \left( {\frac{\pi }{4} + \frac{{k\pi }}{6}} \right)}}} \cr & = \sum\limits_{k = 1}^{13} {\frac{1}{{\sin \frac{\pi }{6}}}\left[ {\frac{{\sin \left\{ {\frac{\pi }{4} + \frac{{k\pi }}{6} - \left( {\frac{\pi }{4} + \left( {k - 1} \right)\frac{\pi }{6}} \right)} \right\}}}{{\sin \left( {\frac{\pi }{4} + \left( {k - 1} \right)\frac{\pi }{6}} \right)\sin \left( {\frac{\pi }{4} + \frac{{k\pi }}{6}} \right)}}} \right]} \sum\limits_{k = 1}^{13} {2\left[ {\cot \left( {\frac{\pi }{4} + \left( {k - 1} \right)\frac{\pi }{6}} \right) - \cot \left( {\frac{\pi }{4} + \frac{{k\pi }}{6}} \right)} \right]} \cr & = 2\left[ {\left\{ {\cot \frac{\pi }{4} - \cot \left( {\frac{\pi }{4} + \frac{\pi }{6}} \right)} \right\} + \left\{ {\cot \left( {\frac{\pi }{4} + \frac{\pi }{6}} \right) - \cot \left( {\frac{\pi }{4} + \frac{{2\pi }}{6}} \right)} \right\} + ..... + \left\{ {\cot \left( {\frac{\pi }{4} + \frac{{12\pi }}{6}} \right) - \cot \left( {\frac{\pi }{4} + \frac{{13\pi }}{6}} \right)} \right\}} \right] \cr & = 2\left[ {\cot \frac{\pi }{4} - \cot \left( {\frac{\pi }{4} + \frac{{13\pi }}{6}} \right)} \right] \cr & = 2\left[ {1 - \cot \frac{{5\pi }}{{12}}} \right] \cr & = 2\left[ {1 - \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right] \cr & = 2\left[ {1 - \left( {2 - \sqrt 3 } \right)} \right] \cr & = 2\left( {\sqrt 3 - 1} \right). \cr} $$

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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