Question

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

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

Answer 1

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

Given $x_1,x_2,x_3,........,x_n$ are in A.P
and, $x_n-x_{n-1}=................=x_3-x_2=x_2-x_1=\alpha$
Now,
$\sin \alpha (\sec x_1.\sec x_2+\sec x_2.\sec x_3+...........+\sec x_{n-1}.\sec x_n)$
$=\frac{\sin (x_2-x_1)}{\cos x_1.\cos x_2}+\frac{\sin (x_3-x_2)}{\cos x_2.\cos x_3}+.......+\frac{\sin (x_n-x_{n-1})}{\cos x_{n-1}.\cos x_n}$
$=\frac{\sin x_2.\cos x_1-\cos x_2.\sin x_1)}{\cos x_1.\cos x_2}+\frac{\sin x_3.\cos x_2-\cos x_3.\sin x_2)}{\cos x_2.\cos x_3}+.......+\frac{\sin x_n.\cos x_{n-1}-\cos x_n.\sin x_{n-1})}{\cos x_{n-1}.\cos x_n}$
$=\frac{\sin x_n}{\cos x_n}-\frac{\sin x_1}{\cos x_1}=\frac{\sin (x_n-x_1)}{\cos x_1.\cos x_n}=\frac{\sin \alpha (n-1)}{\cos x_1.\cos x_n}$
Hence, option 'A' is correct.