Question

8) tana + tanß
cota + cotp
tana tanß​

Answer 1

Step-by-step explanation:

tanA+tanB

=tanAtanB

Let’s consider the LHS,

LHS =\frac{\tan A+\tan B}{\cot A+\cot B}=

cotA+cotB

tanA+tanB

\Rightarrow \frac{\tan A+\tan B}{\frac{1}{\tan A}+\frac{1}{\tan B}}⇒

tanA

1

+

tanB

1

tanA+tanB

\left[{since}, \cot \theta=\frac{1}{\tan \theta}\right][since,cotθ=

tanθ

1

]

\Rightarrow \frac{\tan A+\tan B}{\frac{\tan A+\tan B}{\tan A \tan B}}⇒

tanAtanB

tanA+tanB

tanA+tanB

\Rightarrow \tan A \tan B⇒tanAtanB

= RHS

Hence proved.