The value of -11-4-k-0107 -12-k -2 7 -12-k-1 -2 in the interval -4- 3 -4- equals

^ 1(1/4∑_k=0^101/cos(7π/12+kπ/2)cos(7π/12+(k+1)π/2)) ^ 1(1/4∑_k=0^10sin[(7π/12+(k+1)π/2) (7π/12+kπ/2)]/cos(7π/12+kπ/2)cos(7π/12+(k+1)π/2)) ∵sin (A B)=sin A·c ...

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Standard XII

Mathematics

Property 4

The value of ...

Question

The value of ${sec}^{- 1} (\frac{1}{4} 10 \sum k = 0 sec (\frac{7 π}{12} + \frac{k π}{2}) sec (\frac{7 π}{12} + \frac{(k + 1) π}{2}))$ in the interval $[- \frac{π}{4}, \frac{3 π}{4}]$ equals

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Solution

${sec}^{- 1} ⎛ ⎜ ⎝ \frac{1}{4} 10 \sum k = 0 \frac{1}{cos (\frac{7 π}{12} + \frac{k π}{2}) cos (\frac{7 π}{12} + \frac{(k + 1) π}{2})} ⎞ ⎟ ⎠$

${sec}^{- 1} ⎛ ⎜ ⎝ \frac{1}{4} 10 \sum k = 0 \frac{sin [(\frac{7 π}{12} + \frac{(k + 1) π}{2}) - (\frac{7 π}{12} + \frac{k π}{2})]}{cos (\frac{7 π}{12} + \frac{k π}{2}) cos (\frac{7 π}{12} + \frac{(k + 1) π}{2})} ⎞ ⎟ ⎠$
$∵ sin (A - B) = sin A \cdot cos B - cos A \cdot sin B$

${sec}^{- 1} ⎛ ⎜ ⎝ \frac{1}{4} 10 \sum k = 0 \frac{(sin (\frac{7 π}{12} + \frac{(k + 1) π}{2}) cos (\frac{7 π}{12} + \frac{k π}{2})) - (cos (\frac{7 π}{12} + \frac{(k + 1) π}{2}) sin (\frac{7 π}{12} + \frac{k π}{2}))}{cos (\frac{7 π}{12} + \frac{k π}{2}) cos (\frac{7 π}{12} + \frac{(k + 1) π}{2})} ⎞ ⎟ ⎠$

${sec}^{- 1} (\frac{1}{4} 10 \sum k = 0 (tan (\frac{7 π}{12} + \frac{(k + 1) π}{2})) - (tan (\frac{7 π}{12} + \frac{k π}{2})))$

${sec}^{- 1} \frac{1}{4} (tan ((\frac{7 π}{12} + \frac{π}{2}) - tan (\frac{7 π}{12}) + (tan (\frac{7 π}{12} + \frac{2 π}{2}) - tan (\frac{7 π}{12} + \frac{π}{2}) + . . . . + (tan (\frac{7 π}{12} + \frac{11 π}{2}) - tan (\frac{7 π}{12} + \frac{10 π}{2}))$

${sec}^{- 1} (\frac{1}{4} (tan (\frac{7 π}{12} + \frac{11 π}{2}) - (tan \frac{7 π}{12})))$

${sec}^{- 1} (\frac{1}{4} tan (6 π + \frac{π}{12}) - (tan \frac{7 π}{12}))$

${sec}^{- 1} (\frac{1}{4} (tan \frac{π}{12}) - (tan \frac{7 π}{12}))$

${sec}^{- 1} (\frac{1}{4} ((2 - \sqrt{3}) + (2 + \sqrt{3})))$

${sec}^{- 1} (1) = 0$

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Similar questions

Q. The value of

{sec}^{- 1} (\frac{1}{4} 10 \sum k = 0 sec (\frac{7 π}{12} + \frac{k π}{2}) sec (\frac{7 π}{12} + \frac{(k + 1) π}{2}))

in the interval

[- \frac{π}{4}, \frac{3 π}{4}]

equals

Q. If

{tan}^{- 1} [20 \sum k = 0 sec (\frac{5 π}{12} + \frac{k π}{2}) sec (\frac{5 π}{12} + \frac{(k + 1) π}{2})] = - {tan}^{- 1} a

, then the value of

a

Q. The value of

\sum_{k = 1}^{13} \frac{1}{s i n (\frac{π}{4} + \frac{(k - 1) π)}{6}) s i n (\frac{π}{4} + \frac{k π}{6})}

is equal to

Q. The value of

13 \sum k = 1 \frac{1}{sin (\frac{π}{4} + \frac{(k - 1) π}{6}) sin (\frac{π}{4} + \frac{k π}{6})}

is equal to

Q. The value of

13 \sum k = 1 \frac{1}{sin (\frac{π}{4} + \frac{(k - 1) π}{6}) sin (\frac{π}{4} + \frac{k π}{6})}

is equal to

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Property 4

Standard XII Mathematics

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The value of sec−1(1410∑k=0sec(7π12+kπ2)sec(7π12+(k+1)π2)) in the interval [−π4,3π4] equals

The value of ${sec}^{- 1} (\frac{1}{4} 10 \sum k = 0 sec (\frac{7 π}{12} + \frac{k π}{2}) sec (\frac{7 π}{12} + \frac{(k + 1) π}{2}))$ in the interval $[- \frac{π}{4}, \frac{3 π}{4}]$ equals