Question

The value of $\sum _{r=1}^7{\tan }^2\left(\frac{r\pi }{16}\right)$ is

• A. $29$
• B. $33$
• C. $34$
• D. $35$

Answer 1

Answer: (D)

$35$

The value of $\sum _{r=1}^7$ is

Let $a=\frac{\pi }{16}$

$\tan 7a=\cot a$ since$(\frac{\pi }{2}-\frac{7\pi }{16}=\frac{\pi }{16})$

Similarly, we get, $\tan 6a=\cot 2a$

$\tan 5a=\cot 3a$

and $4a=\frac{\pi }{4}\tan 4a=\perp$

therefore given expression reduces to

$ta{n}^{2}a+co{t}^{2}a+ta{n}^2\left(2a\right)+co{t}^2\left(2a\right)+ta{n}^2\left(3a\right)+c{t}^2\left(3a\right)+1$

$\Rightarrow {\tan }^{2}a+{\cot }^{2}a=\sin a+{\cos }^{4}a/{\sin }^{2}a{\cos }^{2}a$

$=({\sin }^{2}a+{\cos }^a{)}^2-2{\sin }^{2}a{\cos }^{2}a$

$=(1-2{\sin }^{2}a{\cos }^{2}a)/{\sin }^{2}a{\cos }^{2}a$

$=\frac{1}{{\sin }^{2}a{\cos }^{2}a-2}$

$=\left(4/4{\sin }^{2}a{\cos }^{2}a\right)-2=4/{\sin }^2\left(2a\right)-2$

In the same way simplify other sums we get ,

$4\left(1/{\sin }^2\right(2a)+1/{\sin }^2(4a)+1/{\sin }^2(6a\left)\right)-5$

$1/{\sin }^2\left(4a\right)=2$

$=4(\frac{1}{{\sin }^2\left(2a\right)}+\frac{1}{{\sin }^2\left(6a\right)})+3=8(\frac{1}{1-\cos 4a}+\frac{1}{1-\cos 2a})+3$

$[4a=\pi /4\&12a=3\pi /4]$

this gives $=32+3=35$ as, option (D) is correct.