If {x_1}, {x_2}, {x_3}, ..., {x_n} are in A.P | vedclass

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(A) Given that ${x_1}, {x_2}, ..., {x_n}$ are in $A.P.$ with common difference $\alpha$,so ${x_{k+1}} - {x_k} = \alpha$ for all $k$.We can write $\sin

Vedclass · Accepted Answer

$\frac{\sin (n - 1)\alpha}{\cos {x_1}\cos {x_n}}$

Vedclass · Answer

(A) Given that ${x_1}, {x_2}, ..., {x_n}$ are in $A.P.$ with common difference $\alpha$,so ${x_{k+1}} - {x_k} = \alpha$ for all $k$.We can write $\sin \alpha = \sin({x_{k+1}} - {x_k})$.Thus,the expression becomes $\sum_{k=1}^{n-1} \sin \alpha \sec {x_k} \sec {x_{k+1}} = \sum_{k=1}^{n-1} \frac{\sin({x_{k+1}} - {x_k})}{\cos {x_k} \cos {x_{k+1}}}$.Using the identity $	an A - 	an B = \frac{\sin(A - B)}{\cos A \cos B}$,we get $\sum_{k=1}^{n-1} (	an {x_{k+1}} - 	an {x_k})$.This is a telescoping series: $(	an {x_2} - 	an {x_1}) + (	an {x_3} - 	an {x_2}) + ... + (	an {x_n} - 	an {x_{n-1}})$.After cancellation,we are left with $	an {x_n} - 	an {x_1} = \frac{\sin({x_n} - {x_1})}{\cos {x_n} \cos {x_1}}$.Since ${x_n} = {x_1} + (n-1)\alpha$,then ${x_n} - {x_1} = (n-1)\alpha$.Therefore,the value is $\frac{\sin(n - 1)\alpha}{\cos {x_1} \cos {x_n}}$.

If ${x_1}, {x_2}, {x_3}, ..., {x_n}$ are in $A.P.$ whose common difference is $\alpha$,then the value of $\sin \alpha (\sec {x_1}\sec {x_2} + \sec {x_2}\sec {x_3} + ... + \sec {x_{n - 1}}\sec {x_n}) = $

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