The value of -n- 0881cos nk-cos n -1 k- where k is 1o is equal to

The correct option is B cos k cos ec^2k Multiply amp; divide by sin k T_n=1sin k.sin [ ( n +1 )k ( nk ) #160; ]cos ( n+ 1 )k.cos #160 ...

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Sum of Trigonometric Ratios in Terms of Their Product

The value of ...

Question

The value of $\sum_{n = 0}^{88} \frac{1}{c o s n k . c o s (n + 1) k}$ , where k is $1^{o}$ is equal to

\frac{s i n^{2} k}{c o s k}

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\frac{c o s^{2} k}{s i n k}

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c o s k c o s e c^{2} k

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none of these

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K = 1^{\circ}

88 \sum n = 0 \frac{1}{cos n K . cos (n + 1) K} = \frac{cos K}{{sin}^{2} K}

F (k) = (1 + sin \frac{π}{2 k}) (1 + sin (k - 1) \frac{π}{2 k}) (1 + sin (2 k + 1) \frac{π}{2 k}) (1 + sin (3 k - 1) \frac{π}{2 k})

F (1) + F (2) + F (3)

The value of $\sum_{r = 1}^{n + 1} (\sum_{k = 1}^{n}^{k} C_{r - 1})$ where r, k, n $ϵ$ N is equal to

k \in N,

\frac{sin 2 k x}{sin x} = 2 [cos x + cos 3 x + . . . + cos (2 k - 1) x],

I = \int_{0}^{π / 2} sin 2 k x cot x d x

S_{n} = \sum_{k = 0}^{n} (- 1)^{K} .^{3 n} C_{k}, w h e r e n = 12, . . . . . .

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