Question

If cotA-tanB=0,then show that cot A cot B=1(where A and B acute angle)

Answer 1

Answer:

Hence Proved that CotACotB = 1

Step-by-step explanation:

$CotA - TanB = 0\\\\=> \frac{CosA}{SinA} - \frac{SinB}{CosB} = 0\\\\=> \frac{CosA}{SinA} = \frac{SinB}{CosB}\\\\=> \frac{CosA}{SinA} \times \frac{CosB}{SinB} = 1\\\\=> CotA \times CotB = 1 (as\ \frac{CosA}{SinA} = CotA)$

Hence Proved that CotACotB = 1

Answer 2

Answer:

The proof is explained step-wise below :

Step-by-step explanation:

Given that : cot A - tan B = 0

To Prove : cot A × cot B = 1

Proof :

$\cot A - \tan B = 0\\\\\implies \frac{\cos A}{\sin A}-\frac{\sin B}{\cos B}=0\\\\ \implies \frac{\cos A}{\sin A}=\frac{\sin B}{\cos B}\\\\\implies \frac{\cos A}{\sin A} \times \frac{\cos B}{\sin B}=1\\\\\implies \cot A\times\cot B = 1$

Hence Proved.