Question

$cot(45°+\theta )cot(45°-\theta )=$

• A. $-1$
• B. $0$
• C. $1$
• D. $\infty$

Answer 1

The correct option is C

$1$

Explanation for the correct option:

Step 1. Find the value of given expression:

Given, $cot(45°+\theta )cot(45°-\theta )$ …….$\left(1\right)$

As we know,

$\mathit{c}\mathit{o}\mathit{t}\mathbf{(}\mathit{A}\mathbf{+}\mathit{B}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{(}\mathbf{C}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{A}\mathbf{.}\mathbf{C}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{B}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{(}\mathbf{C}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{A}\mathbf{}\mathbf{+}\mathbf{}\mathbf{C}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{B}\mathbf{)}\mathbf{}}$

$\mathit{c}\mathit{o}\mathit{t}\mathbf{(}\mathit{A}\mathbf{-}\mathit{B}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{(}\mathbf{C}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{A}\mathbf{.}\mathbf{C}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{B}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{(}\mathbf{C}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{B}\mathbf{-}\mathbf{C}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{A}\mathbf{)}}$

Step 2. Multiply both formula, we get

$\Rightarrow$$cot(A+B)cot(A-B)=\frac{(CotA.CotB-1)\times (CotA.CotB+1)}{(CotB-CotA)\times (CotB+CotA)}$

$\Rightarrow$$cot(A+B)cot(A-B)=\frac{(CotA.CotB)²-1}{(CotB²-CotA²)}$ …….(2)

Comparing equation $\left(1\right)$ and $\left(2\right)$, we get

$A=45,B=\theta$

Step 3. Put these values in equation $\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

$\Rightarrow$$cot(45°+\theta )cot(45°-\theta )=\frac{(Cot45°.Cot\theta )²-1)}{(Cot\theta ²-Cot45²)}$

$=\frac{cot²\theta -1}{cot²\theta -1}$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{t}\mathbf{}\mathbf{45}\mathbf{°}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$=1$

Hence, Option ‘C’ is Correct.