Question

The value of $\tan 67{\frac{1}{2}}^{\circ }+\cot 67{\frac{1}{2}}^{\circ }$ is.

• A. $\sqrt{2}$
• B. $3\sqrt{2}$
• C. $2\sqrt{2}$
• D. $2-\sqrt{2}$

Answer 1

Answer: (C)

$2\sqrt{2}$

Let ${67.5}^o=x$

$\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}$

$\tan 2x=\tan \left({135}^0\right)=\tan ({90}^o+{45}^o)=-\cot \left({45}^0\right)=-1$

So, $-1=\frac{2\tan x}{1-{\tan }^{2}x}$

$\Rightarrow -(1-{\tan }^{2}x)=2\tan x\Rightarrow {\tan }^{2}x-2\tan x-1=0\Rightarrow \tan x=\frac{-(-2)\pm \sqrt{{(-2)}^2-4(-1)}}{2}$

$\Rightarrow \tan x=\frac{2\pm 2\sqrt{2}}{2}=1+\sqrt{2}$

Now, $\cot x=\frac{1}{\tan x}=\frac{1}{1+\sqrt{2}}\times \frac{1-\sqrt{2}}{1-\sqrt{2}}=\sqrt{2}-1$

$\therefore \tan {67.5}^0+\cot {67.5}^0=\tan x+cotx=1+\sqrt{2}+\sqrt{2}-1=2\sqrt{2}$