Question

If θ1, θ2, θ3........,θn are in A.P., whose common difference is d, then,
sec θ1 sec θ2+sec θ2 sec θ3+..........+sec θn−1 sec θn=

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

Answer 1

The correct option is A $\frac{\tan {\theta }_n-\tan {\theta }_1}{\sin d}$

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