Question

prove that cot A+tanB/cotB+tanA = cotA tanB

Answer 1

LHS: cotA + tanB / cotB + tanA

= (cotA + 1/cotB) / (cotB + 1/cotA)

 = (cotAcotB + 1) / cotB / (cotAcotB + 1) / cotA

= cotA / cotB

= cotAtanB

= RHS

hence proved.

Answer 2

$\bold{\frac{(\cot A+\tan B)}{(\cot B+\tan A)}=\cot A . \tan B}$

Given:

$\frac{(\cot A+\tan B)}{(\cot B+\tan A)}=\cot A \cdot \tan B$

To Prove:

L.H.S = R.H.S

Proof:

Let simplify the question in form sin and cos

$\frac{(\cot A+\tan B)}{(\cot B+\tan A)}=\cot A \cdot \tan B$

L.H.S

$\frac{(\cot A+\tan B)}{(\cot B+\tan A)}$

Simplifying it in terms of sin and cos take the LCM of the denominator  

$=\frac{\left(\frac{\cos A}{\sin A}+\frac{\sin B}{\cos B}\right)}{\left(\frac{\cos B}{\sin B}+\frac{\sin A}{\cos A}\right)}$

Deducting the value of $\cos A \cos B+\sin A \sin B$ on both sides, we get:

$=\frac{\frac{\cos A \cos B+\sin A \sin B}{\sin A \cos B}}{\frac{\cos B \cos A+\sin B \sin A}{\sin B \cos A}}$

After crossing out,  $\cos B \cos A+\sin B \sin A$  we get the value of  

$\frac{\cos A \cos B+\sin A \sin B}{\cos B \cos A+\sin B \sin A}, \quad \frac{\sin B \cos A}{\sin A \cos B}$

Cancelling out and transforming the identities into cot and tan,  

$\bold{\frac{\cos A \sin B}{\sin A \cos B}=\frac{\cos A}{\sin A} \cdot \frac{\sin \mathrm{B}}{\cos B}=\cot A \tan B=R . H . S}$

Hence L.H.S = R. H.S

Hence Proved

Therefore, it proves that $\bold{\frac{(\cot A+\tan B)}{(\cot B+\tan A)}=\cot A \cdot \tan B}$.