Question

(tanA+tanB)cosAcosB=sin (A+B)

Answer 1

Here we will use the following concepts:

$\boxed{\sin A \cos B + \cos A \sin B = \sin (A+B)} \\ \\ \\ \boxed{\tan \theta = \frac{\sin \theta}{\cos \theta}}$

Now, we have to prove:

$(\tan A + \tan B) \cos A \cos B = \sin (A+B)$

We can prove it as follows:

$\mathbb{LHS} \\ \\ \\ = (\tan A +\tan B) \cos A \cos B \\ \\ \\ = \tan A \cos A \cos B + \tan B \cos A \cos B \\ \\ \\ = \frac{\sin A}{\cancel{\cos A}} \, \cancel{\cos A} \cos B + \frac{\sin B}{\cancel{\cos B}} \cos A \, \cancel{\cos B} \\ \\ \\ = \sin A \cos B + \cos A \sin B \\ \\ \\ = \sin (A+B) \\ \\ \\ = \mathbb{RHS}$


Hope it helps
Purva
Brainly Community

Answer 2

taking LHS:

tanA.cosA.cosB+tanB.cosA.cosB       .'. (tanA=sinA/CosA and tanB=sinB/cosB)

sinA.cosB+sinB.cosA

sin(A+B)