Question

If $\alpha +\beta =\frac{5\pi }{4}$, then value of $\frac{\cot \alpha \cot \beta }{(1+\cot \alpha )(1+\cot \beta )}$ is equal to

• A. $1$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $\frac{5}{4}$

Answer 1

The correct option is C $\frac{1}{2}$

If $\alpha +\beta =\frac{5\pi }{4}$

Taking both side ${}^{\prime }{\cot }^{\prime }$ and we get

$\cot (\alpha +\beta )=\cot \frac{5\pi }{4}$

$\frac{\cot \alpha \cot \beta -1}{\cot \beta +\cos \alpha }=\cot (\pi +\frac{\pi }{4})$

$\therefore \cot (\alpha +\beta )=\frac{\cot \alpha \cot \beta -1}{\cot \beta +\cot \alpha }$

$\Rightarrow \frac{\cot \alpha \cot \beta -1}{\cot \beta +\cot \alpha }=\cot \frac{\pi }{4}$ $\therefore \cot (\pi +\theta )=\cot \theta$

$\Rightarrow \frac{\cot \alpha \cot \beta -1}{\cot \beta +\cot \alpha }=1$ $\therefore \cot \frac{\pi }{4}=1$

$\Rightarrow \cot \alpha \cot \beta -1=\cot \beta +\cot \alpha$

on adding both side$(\cot \alpha \cot \beta +1)$

$\Rightarrow \cot \alpha \cot \beta -1+\cot \alpha \cot \beta +1=\cot \beta +\cot \alpha +\cot \alpha \cot \beta +1$

$\Rightarrow 2\cot \alpha \cot \beta =\cot \beta +\cot \alpha \cot \beta +\cot \alpha +1$

$\Rightarrow 2\cot \alpha \cot \beta =\cot \beta (+\cot \alpha )+1(1+\cot \alpha )$

$\Rightarrow 2\cot \alpha \cot \beta =(1+\cot \alpha )(1+\cot \beta )$

$\Rightarrow (1+\cot \alpha )(1+\cot \beta )=2\cot \alpha \cot \beta$

$\Rightarrow \frac{(1+\cot \alpha )(1+\cot \beta )}{\cot \alpha \cot \beta )}=2$

On reciprocal that

$\Rightarrow \frac{\cot \alpha \cot \beta }{(1+\cot \alpha )(1+\cot \beta )}=\frac{1}{2}$

Hence, this is the answer.