Question

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

• A. $\frac{\sin \theta }{\cos x_{1}cos{x}_n}$
• 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 $\frac{\sin (n-1)\theta }{\cos x_1\cos x_n}$

Given $x_1,x_2,x_3,........,x_n$ are in A.P

and, $x_n-x_{n-1}=................=x_3-x_2=x_2-x_1=\theta$ $\dots \left(1\right)$

Now,
$sin\theta (sec{x}_1.sec{x}_2+sec{x}_2.sec{x}_3+...........+sec{x}_{n-1}.sec{x}_n)$

$=\frac{sin\left(\theta \right)}{cos{x}_1.cos{x}_2}+\frac{sin\left(\theta \right)}{cos{x}_2.cos{x}_3}+.......+\frac{sin\left(\theta \right)}{cos{x}_{n-1}.cos{x}_n}$

$=\frac{sin(x_2-x_1)}{cos{x}_1.cos{x}_2}+\frac{sin(x_3-x_2)}{cos{x}_2.cos{x}_3}+.......+\frac{sin(x_n-x_{n-1})}{cos{x}_{n-1}.cos{x}_n}$ from $\left(1\right)$
$=\frac{sin{x}_2.cos{x}_1-cos{x}_2.sin{x}_1}{cos{x}_1.cos{x}_2}+\frac{sin{x}_3.cos{x}_2-cos{x}_3.sin{x}_2}{cos{x}_2.cos{x}_3}+.......+\frac{sin{x}_n.cos{x}_{n-1}-cos{x}_n.sin{x}_{n-1}}{cos{x}_{n-1}.cos{x}_n}$

$=\frac{sin{x}_2}{cos{x}_2}-\frac{sin{x}_1}{cos{x}_1}+\frac{sin{x}_3}{cos{x}_3}-\frac{sin{x}_2}{cos{x}_2}+\frac{sin{x}_4}{cos{x}_4}-\frac{sin{x}_3}{cos{x}_3}+............+\frac{sin{x}_n}{cos{x}_n}-\frac{sin{x}_{(n-1)}}{cos{x}_{(n-1)}}$
$=\frac{sin{x}_n}{cos{x}_n}-\frac{sin{x}_1}{cos{x}_1}=\frac{sin(x_n-x_1)}{cos{x}_1.cos{x}_n}=\frac{sin\theta (n-1)}{cos{x}_1.cos{x}_n}$

Hence, option 'B' is correct.