Question

If ${\theta }_1,{\theta }_2,{\theta }_3,\dots ,{\theta }_n$ are in AP, whose common difference is $d$, then $\sin d(\sec {\theta }_1\sec {\theta }_2+\sec {\theta }_2\sin {\theta }_3+\cdots +\sec {\theta }_{n-1}\sec {\theta }_n)$ is equal to

• A. $\tan {\theta }_n-\tan {\theta }_2$
• B. $\tan {\theta }_n+\tan {\theta }_1$
• C. $\tan {\theta }_n-\tan {\theta }_1$
• D. None of these

Answer 1

The correct option is C $\tan {\theta }_n-\tan {\theta }_1$

Given, ${\theta }_1,{\theta }_2,{\theta }_3,\dots ,{\theta }_n$ are in AP.
$\Rightarrow {\theta }_2-{\theta }_1={\theta }_3-{\theta }_2=\cdots ={\theta }_n-{\theta }_{n-1}=d$ ...(i)
Now, taking only first term,
$\sin d\sec {\theta }_1\sec {\theta }_2=\frac{\sin d}{\cos {\theta }_1\cos {\theta }_2}$
$=\frac{\sin ({\theta }_2-{\theta }_1)}{\cos {\theta }_1\cos {\theta }_2}$ [from equation (i)]
$=\frac{\sin {\theta }_2\cos {\theta }_1-\cos {\theta }_2\sin {\theta }_1}{\cos {\theta }_1\cos {\theta }_2}$
$=\frac{\sin {\theta }_2\cos {\theta }_1}{\cos {\theta }_1\cos {\theta }_2}-\frac{\cos {\theta }_2\sin {\theta }_1}{\cos {\theta }_1\cos {\theta }_2}$
$=\tan {\theta }_2-\tan {\theta }_1$
Similarly, we can solve other terms which will be
$\tan {\theta }_3-\tan {\theta }_2$, $\tan {\theta }_4-\tan {\theta }_3,\dots$.
$\therefore \sin d(\sec {\theta }_1\sec {\theta }_2+\sec {\theta }_2\sec {\theta }_3+\cdots +\sec {\theta }_{n-1}\sec {\theta }_n)$
$=\tan {\theta }_2-\tan {\theta }_1+\tan {\theta }_3-{\theta }_2+\cdots +\tan {\theta }_n-\tan {\theta }_{n-1}$
$=-\tan {\theta }_1+\tan {\theta }_n$
$=\tan {\theta }_n-\tan {\theta }_1$