Question

please solve the question tan A + B by cot a minus b equal to tan square A minus tan square b by 1 minus tan square into tan square B please solve my question very quickly .I need to help​ .​

Answer 1

Solution:

Proper Question:

Tan(A+B)/Cot(A-B)= tan²A - tan²B/1-tan²A.tan²B

Explanation:

The given expression is Tan(A+B)/Cot(A-B).

We will use some Trigonometric Compound Identities to solve this question:

Explanation:

Tan(A+B) = (Tan A + Tan B/ 1 - TanA.TanB)

Tan(A-B) = (Tan A - Tan B/ 1 + tanA.tanB

Explanation:

Tan(A+B)/Cot(A-B)

Tan(A+B) × tan(A-B)

(Tan A + Tan B/ 1 - TanA.TanB) × Tan A - Tan B/ 1 + tanA.tanB

Tan²A - Tan²B = 1 - tan²A.Tan²B

Answer 2

$\huge\sf{Answer:-}$

Given

$\sf =\frac{Tan(A+B)}{Cot(A-B)}$

Trigonometric Compound Identities

Hence,

$\sf =Tan \: (A \: + \: B) = \frac{(Tan \: A \: + \: Tan \: B)}{(1 \: - Tan \: * \: Tan \: B)}$

$\sf =Tan(A-B) = \frac{Tan \: A \: - \: Tan \: B}{1 \: + \: tan \: A \: tan \: B}$

So,

$\sf = \frac{Tan \: (A \: + \: B)}{Cot \: (A \: - \: B)}$

$\sf = Tan \: (A \: + \: B) × tan \: (A \: - \: B)$

$\sf =\frac{(Tan A + Tan B}{1 - (TanA \: TanB) } × \frac{Tan A - Tan B}{1 + tanA*tanB}$

= Tan²A - Tan²B = 1 - tan²A*Tan²B