Question

If $x_1,x_2,x_3,\dots x_n$ in AP, whose common difference is $\theta$, then the value of $\sin \theta (\sec x_1\sec x_2+\sec x_2\sec x_3+\dots \dots ..+\sec x_{n-1}\sec x_n)$ is

• A. $\frac{\sin n\theta }{\cos x_1\cos x_n}$
• B. $\frac{\sin (n-1)\theta }{\cos x_1\cos x_n}$
• C. $\sin n\theta \cos x_1\cos x_n$
• D. $\frac{\cos (n-1)\theta }{\cos x_1\cos x_n}$

Answer 1

The correct option is B $\frac{\sin (n-1)\theta }{\cos x_1\cos x_n}$

To find:

$\sin \theta (\sec x_1\sec x_2+\sec x_2\sec x_3+\cdots +\sec x_{n-1}\sec x_n$)

It is given that $x_1,x_2,............x_n$ are in A.P

Whose common difference is $\theta$

Since $\theta$ is the common difference
Therefore ,
$x_2-x_1=\theta$
$x_3-x_2=\theta$

.

.

$x_n-x_{n-1}=\theta$
Now,
$\sin \theta (\sec x_1\sec x_2+\sec x_2\sec x_3+\cdots +\sec x_{n-1}\sec x_n)$
This can be written as
$=\sin \theta (\frac{1}{\cos x_1\cos x_2}+\frac{1}{\cos x_2\cos x_3}+\cdots \frac{1}{\cos x_{n-1}\cos x_n})$

$=(\frac{\sin \theta }{\cos x_1\cos x_2}+\frac{\sin \theta }{\cos x_2\cos x_3}+\cdots \frac{\sin \theta }{\cos x_{n-1}\cos x_n})$
From using above equations, we get
$=(\frac{\sin (x_2-x_1)}{\cos x_1\cos x_2}+\frac{\sin (x_3-x_2)}{\cos x_2\cos x_3}+\cdots \frac{\sin (x_n-x_{n-1})}{\cos x_{n-1}\cos x_n})$

As we know that,

$\tan A-\tan B=\frac{\sin (A-B)}{\cos A\cos B}$

$\therefore$ On using this, we get
$=(\frac{\sin (x_2-x_1)}{\cos x_1\cos x_2}+\frac{\sin (x_3-x_2)}{\cos x_3\cos x_2}+\cdots \frac{\sin (x_n-x_{n-1})}{\cos x_n\cos x_{n-1}})$

$=\tan x_2-\tan x_1+\tan x_3-\tan x_2+\cdots +\tan x_n-\tan x_{n-1}$
Finally, we get
$=\tan x_n-\tan x_1$
$=\left(\frac{\sin (x_n-x_1)}{\cos x_n\cos x_1}\right)$
Also ,
$x_n$ can be written as
$x_n=(n-1)\theta +x_1$
On putting this value, we get

$=\left(\frac{\sin \left(\right(n-1)\theta +x_1-x_1)}{\cos x_n\cos x_1}\right)$

$=\left(\frac{\sin (n-1)\theta }{\cos x_n\cos x_1}\right)$

Hence, option B.