Question

$\cot {16}^o\cdot \cot {44}^o+\cot {44}^o\cdot {76}^o-\cot {76}^o.\cot {16}^o=$

• A. $3$
• B. $0$
• C. $1$
• D. $4$

Answer 1

Answer: (A)

$3$

$\cot 16\cot 44+\cot 44\cot 76-\cot 76\cot 16$

Here $16+44=60,44+76=12\&76-16=60$

we know $\cot (A+B)=\frac{\cot A.\cot B-1}{\cot A+\cot B}$

$\cot (A-B)=\frac{\cot A\cot B+1}{\cot B-\cot A}$

$\cot 60=\cot (16+44)$

$\frac{1}{\sqrt{3}}=\frac{\cot 16\cot 44-1}{\cot 16+\cot 44}$

$1+\frac{\cot 16+\cot 44}{\sqrt{3}}=\cot 16\cot 44$ ..$\left(1\right)$

$\frac{1+\cot 16+\cot 44}{-\sqrt{3}}=\cot 44\cot 76$ ..$\left(2\right)$

$-1+\frac{\cot 16-\cot 76}{\sqrt{3}}=\cot 76\cot 16$ ..$\left(3\right)$

$\left(1\right)+\left(2\right)-\left(3\right)$ gives us.

$\cot 16\cot 44+\cot 44\cot 76-\cot 76\cot 16$

$=\frac{\cot 16\cot 44}{\sqrt{3}}+1-\frac{\cot 44\cot 76}{\sqrt{3}}+1-\frac{\cot 16-\cot 76}{\sqrt{3}}+1$

$=3$