The value of -11-4-k-0107-12-k-27-12-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· ...

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Standard XII

Mathematics

Integration of Trigonometric Functions

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

${sec}^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{1}{4} 10 \sum k = 0 \frac{1}{cos (\frac{7 π}{12} + \frac{k π}{2}) cos (\frac{7 π}{12} + \frac{(k + 1) π}{2})} ⎞ ⎟ ⎟ ⎠$
$= {sec}^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{1}{4} 10 \sum k = 0 \frac{sin [(\frac{7 π}{12} + \frac{(k + 1) π}{2}) - (\frac{7 π}{12} + \frac{k π}{2})]}{cos (\frac{7 π}{12} + \frac{k π}{2}) cos (\frac{7 π}{12} + \frac{(k + 1) π}{2})} ⎞ ⎟ ⎟ ⎠$
$(∵ sin (A - B) = sin A \cdot cos B - cos A \cdot sin B)$
$= {sec}^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{1}{4} 10 \sum k = 0 \frac{(sin (\frac{7 π}{12} + \frac{(k + 1) π}{2}) cos (\frac{7 π}{12} + \frac{k π}{2})) - (cos (\frac{7 π}{12} + \frac{(k + 1) π}{2}) sin (\frac{7 π}{12} + \frac{k π}{2}))}{cos (\frac{7 π}{12} + \frac{k π}{2}) cos (\frac{7 π}{12} + \frac{(k + 1) π}{2})} ⎞ ⎟ ⎟ ⎠$
$= {sec}^{- 1} (\frac{1}{4} 10 \sum k = 0 (tan (\frac{7 π}{12} + \frac{(k + 1) π}{2})) - (tan (\frac{7 π}{12} + \frac{k π}{2})))$
$= {sec}^{- 1} \frac{1}{4} (tan (\frac{7 π}{12} + \frac{π}{2}) - tan (\frac{7 π}{12}) + tan (\frac{7 π}{12} + \frac{2 π}{2}) - tan (\frac{7 π}{12} + \frac{π}{2}) + . . . . + tan (\frac{7 π}{12} + \frac{11 π}{2}) - tan (\frac{7 π}{12} + \frac{10 π}{2}))$
$= {sec}^{- 1} \frac{1}{4} (tan (\frac{7 π}{12} + \frac{11 π}{2}) - tan \frac{7 π}{12})$
$= {sec}^{- 1} \frac{1}{4} (tan (6 π + \frac{π}{12}) - tan (π - \frac{5 π}{12}))$
$= {sec}^{- 1} \frac{1}{4} (tan \frac{π}{12} + tan \frac{5 π}{12})$
$= {sec}^{- 1} \frac{1}{4} [(2 - \sqrt{3}) + (2 + \sqrt{3})]$
$= {sec}^{- 1} (1) = 0$

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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}))

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Integration of Trigonometric Functions

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