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. $\frac{\left[\sin \right(n–1\left)\alpha \right]}{(\cos x_1\cos x_n)}$
• B. $\frac{[\sin n\alpha ]}{(\cos x_1\cos x_n)}$
• C. $\left[\sin \right(n–1\left)\alpha \right](\cos x_1\cos x_n)$
• D. $[\sin n\alpha ](\cos x_1\cos x_n)$

Answer 1

The correct option is A

$\frac{\left[\sin \right(n–1\left)\alpha \right]}{(\cos x_1\cos x_n)}$

Explanation for the correct option :

It is given that, $x_1,x_2,x_3,...,x_n$ are in A.P with $d=\alpha$.

$\Rightarrow x_2-x_1=x_3-x_2=......=x_n-x_{n-1}=\alpha$

$$\begin{gathered} \sin \alpha (sec{x}_{1}sec{x}_2+sec{x}_{2}sec{x}_3+...+sec{x}_{n-1}sec{x}_n) \\ =\frac{\sin \left(x_2-x_1\right)}{\cos x_1\cos x_2}+\frac{\sin \left(x_3-x_2\right)}{\cos x_2\cos x_3}+....+\frac{\sin \left(x_n-x_{n-1}\right)}{\cos x_{n-1}\cos x_n} \\ \mathrm{Using}\sin (\mathrm{A}-\mathrm{B})=\mathrm{sinAcosB}-\mathrm{sinBcosA} \\ =\frac{\sin x_2\cos x_1-\sin x_1\cos x_2}{\cos x_1\cos x_2}+\frac{\sin x_3\cos x_2-\sin x_2\cos x_3}{\cos x_2\cos x_3}+.....+\frac{\sin x_n\cos x_{n-1}-\sin x_{n-1}\cos x_n}{\cos x_{n-1}\cos x_n} \\ =\tan x_2-\tan x_1+\tan x_3-\tan x_2+......+\tan x_n-\tan x_{n-1} \\ =\tan x_n-\tan x_1 \\ =\frac{\sin x_n}{\cos x_n}-\frac{\sin x_1}{\cos x_1} \\ =\frac{\sin x_n\cos x_1-\sin x_1\cos x_n}{\cos x_n\cos x_1} \\ =\frac{\sin \left(x_n-x_1\right)}{\cos x_n\cos x_1} \\ =\frac{\sin \left(n-1\right)\alpha }{\cos x_n\cos x_1} \end{gathered}$$

Hence, option A is correct.