Question

If $A+B=45°$ , then $(cotA-1)(cotB-1)$ is equal to

• A. $3$
• B. $\frac{1}{2}$
• C. $-1$
• D. $-2$
• E. $2$

Answer 1

The correct option is E

$2$

Explanation for the correct option:

Step 1. Find the value of $(cotA-1)(cotB-1)$:

Given, $A+B=45°$

By Taking “cot” on both sides, we get

$cot(A+B)=cot45°$

$\Rightarrow$ $\frac{(cotAcotB–1)}{(cotA+cotB)}=cot45°$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{t}\mathbf{}\left(\theta +\varphi \right)\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{\varphi }\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}}{\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{\varphi }}\right]$

$\Rightarrow$ $cotAcotB–1=cotA+cotB$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{t}\mathbf{45}\mathbf{°}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\Rightarrow$$cotA+cotB–cotAcotB=-1$

Step 2. Subtract 1 from both sides:

$\Rightarrow$$cotA+cotB–cotAcotB-1=-1-1$

$\Rightarrow$ $cotA(1–cotB)–1(1–cotB)=-2$

$\Rightarrow$ $(cotA–1)(1–cotB)=-2$

$\mathbf{\therefore }\mathbf{(}\mathit{c}\mathit{o}\mathit{t}\mathbf{}\mathit{A}\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{(}\mathit{c}\mathit{o}\mathit{t}\mathbf{}\mathit{B}\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}$

Hence, option ‘E’ is Correct.