Question

If ${\theta }_1,{\theta }_2,{\theta }_3,....,{\theta }_n$ are in A.P. whose common difference is $d$, then $\sin d[\sec {\theta }_1\sec {\theta }_2+\sec {\theta }_2\sec {\theta }_3+.....+\sec {\theta }_{n-1}\sec {\theta }_n]=$

• A. $\sec {\theta }_n-\sec {\theta }_1$
• B. $\sin {\theta }_n-\sin {\theta }_1$
• C. $\cos {\theta }_n-\cos {\theta }_1$
• D. $\tan {\theta }_n-\tan {\theta }_1$

Answer 1

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

${\theta }_1,{\theta }_2,{\theta }_3,...{\theta }_n$ are in AP $\Rightarrow d={\theta }_2-{\theta }_1={\theta }_3-{\theta }_2=.....={\theta }_n-{\theta }_{n-1}$

$\left(\sin d\right)[\sec {\theta }_1\sec {\theta }_2+\sec {\theta }_2\sec {\theta }_3+.....\sec {\theta }_{n-1}\sec {\theta }_n]$

$=\frac{\sin d}{\cos {\theta }_1.\cos {\theta }_2}+\frac{\sin d}{\cos {\theta }_2.\cos {\theta }_3}+.....\frac{\sin d}{\cos {\theta }_{n-1}.\cos {\theta }_n}$

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

$=\frac{\sin {\theta }_2.\cos {\theta }_1-\sin {\theta }_1.\cos {\theta }_2}{\cos {\theta }_1.\cos {\theta }_2}+\frac{\sin {\theta }_3.\cos {\theta }_2-\sin {\theta }_2.\cos {\theta }_3}{\cos {\theta }_2.\cos {\theta }_3}+.....\frac{\sin {\theta }_n.\cos {\theta }_{n-1}-\sin {\theta }_{n-1}.\cos {\theta }_n}{\cos {\theta }_{n-1}.\cos {\theta }_n}$

$=[\frac{\sin {\theta }_2}{\cos {\theta }_2}-\frac{\sin {\theta }_1}{\cos {\theta }_1}]+[\frac{\sin {\theta }_3}{\cos {\theta }_3}-\frac{\sin {\theta }_2}{\cos {\theta }_2}]+.....[\frac{\sin {\theta }_n}{\cos {\theta }_n}-\frac{\sin {\theta }_{n-1}}{\cos {\theta }_{n-1}}]$

$=\frac{\sin {\theta }_n}{\cos {\theta }_n}-\frac{\sin {\theta }_1}{\cos {\theta }_1}$

$=\tan {\theta }_n-\tan {\theta }_1$.

The correct answer is option D.