Question

If $\text{A + B = 225°}$, then $\left[\frac{cotA}{\mathrm{1\; +}cotA}\right]\left[\frac{cotB}{\mathrm{1\; +}cotB}\right]=$

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is D

$\frac{1}{2}$

Explanation for the correct option :

Step $\mathbf{1}$ : Expressing the given data:

Given $\text{A + B = 225°}$

Taking " cot " on both sides, $\text{cot(A + B) = cot 225°}$

$\Rightarrow$$\frac{\cot \mathrm{A}\cot \mathrm{B}–1}{\cot \mathrm{A}+\cot \mathrm{B}}=\; 1$ $\left[\because \cot (180°+45°)=\cot 45°=1\right]$

$\Rightarrow$ $\text{cot A cot B – 1 = cot A + cot B}$

$\Rightarrow$ $\text{cot A cot B = 1 + cot A + cot B…..(i)}$

Step $\mathbf{2}$: Simplification of $\left[\frac{cotA}{1+cotA}\right]\left[\frac{cotB}{1+cotB}\right]$

Now, $\left[\frac{cotA}{\mathrm{1\; +}cotA}\right]\left[\frac{cotB}{\mathrm{1\; +}cotB}\right]=\frac{cotAcotB}{(1+cotB)(1+cotA)}$

$=\frac{\cot \mathrm{A}\cot \mathrm{B}}{\mathrm{1\; +}\cot \mathrm{A}+\cot \mathrm{B}+\cot \mathrm{A}\cot \mathrm{B}}$

$=\frac{\cot \mathrm{A}\cot \mathrm{B}}{\cot \mathrm{A}\cot \mathrm{B}+\cot \mathrm{A}\cot \mathrm{B}}$

$=\frac{\cot \mathrm{A}\cot \mathrm{B}}{2\cot \mathrm{A}\cot \mathrm{B}}$

$\left[\frac{cotA}{1+cotA}\right]\left[\frac{cotB}{1+cotB}\right]$$\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$

Hence, option $\mathbf{\text{"D"}}$ is correct.