Question

If $f\left(A\right)=\sum _{r=1}^8\tan rA\cdot \tan (r+1)A$ then

• A. $f\left({18}^{\circ }\right)=-10$
• B. $f\left({36}^{\circ }\right)=-10$
• C. $f\left(\frac{\pi }{4}\right)=-8$
• D. $f\left(\frac{\pi }{8}\right)=-8$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $f\left({18}^{\circ }\right)=-10$
B $f\left({36}^{\circ }\right)=-10$
C $f\left(\frac{\pi }{4}\right)=-8$
D $f\left(\frac{\pi }{8}\right)=-8$

Given : $f\left(A\right)=\sum _{r=1}^8\tan rA\cdot \tan (r+1)A$
We know that
$\tan (r+1)A=\frac{\tan rA+\tan A}{1-\tan rA\tan A}\Rightarrow \tan (r+1)A-\tan (r+1)A\tan rA\tan A=\tan rA+\tan A\Rightarrow \tan (r+1)A\tan rA\tan A=\tan (r+1)A-\tan rA-\tan A$

Now,
$f\left(A\right)=\sum _{r=1}^8\left(\tan \right(r+1)A\cdot \tan rA\cdot \tan A)\cot A\Rightarrow f\left(A\right)=\sum _{r=1}^8\left(\tan \right(r+1)A-\tan rA-\tan A)\cot A\Rightarrow f\left(A\right)=\cot A\left(\sum _{r=1}^8\tan (r+1)A-\tan rA\right)-\sum _{r=1}^{8}1\Rightarrow f\left(A\right)=\cot A(\tan 9A-\tan A)-8\Rightarrow f\left(A\right)=\cot A\tan 9A-9$

Therefore,
$f\left({18}^{\circ }\right)=\tan (9\times {18}^{\circ })\cot {18}^{\circ }-9\Rightarrow f\left({18}^{\circ }\right)=-\tan \left({18}^{\circ }\right)\cot {18}^{\circ }-9\Rightarrow f\left({18}^{\circ }\right)=-10$

$f\left({36}^{\circ }\right)=\tan (9\times {36}^{\circ })\cot {36}^{\circ }-9\Rightarrow f\left({36}^{\circ }\right)=-\tan \left({36}^{\circ }\right)\cot {36}^{\circ }-9\Rightarrow f\left({36}^{\circ }\right)=-10$

$f\left(\frac{\pi }{4}\right)=\cot \frac{\pi }{4}\tan \frac{9\pi }{4}-9\Rightarrow f\left(\frac{\pi }{4}\right)=-8$

$f\left(\frac{\pi }{8}\right)=\cot \frac{\pi }{8}\tan \frac{9\pi }{8}-9\Rightarrow f\left(\frac{\pi }{8}\right)=-8$