Question

What is $\frac{\cot {224}^o-\cot {134}^o}{\cot {226}^o+\cot {316}^o}$ equal to ?

• A. $-\text{cosec}{88}^o$
• B. $-\text{cosec}{2}^o$
• C. $-\text{cosec}{44}^o$
• D. $-\text{cosec}{46}^o$

Answer 1

The correct option is B $-\text{cosec}{2}^o$

We can write, $\cot \left({224}^o\right)=\cot \left({44}^o\right)$

$\cot \left({134}^o\right)=-\cot \left({46}^o\right)$

$\cot \left({226}^o\right)=\cot \left({46}^o\right)$

$\cot \left({316}^o\right)=-\cot \left({44}^o\right)$

We can rewrite the equation as

$\frac{\cot \left({224}^o\right)-\cot \left({134}^o\right)}{\cot \left({226}^o\right)+\cot \left({316}^o\right)}$

$=\frac{\cot \left({44}^o\right)+\cot \left({46}^o\right)}{\cot \left({46}^o\right)-\cot \left({44}^o\right)}$

Now, $\cot A-\cot B=\frac{\cos A}{\sin A}-\frac{\cos B}{\sin B}$

$\cot A-\cot B=\frac{\sin (B-A)}{\sin A\sin B}$

Similarly,

$\cot A+\cot B=\frac{\sin (B+A)}{\sin A\sin B}$

Putting $A={46}^o$ and $B={44}^o$

we get

$\frac{\cot \left({44}^0\right)+\cot \left({46}^0\right)}{\cot \left({46}^0\right)-\cot \left({44}^0\right)}$

$=\frac{\sin \left({90}^o\right)}{\sin \left({-2}^o\right)}$

$=-\text{cosec}{2}^o$