tanA+tanB/cotA+cotB=tanA+tanB Related: Trigonometric Ratios- 1?

Question

tanA+tanB/cotA+cotB=tanA+tanB Related:  Trigonometric Ratios- 1?

Answer

Check your question it must be ,
tan A+tanB/CotA+CotB=tanA.tanB,
Now see the solution,
LHS:
(tan A+tan B) / (cot A+cot B)
⇒(tan A+tan B) / (1/tan A+1/tan B)
⇒(tan A +tan B) / (tan A+tan B / tan A·tan B)
⇒(tan A+tan B) / (tan A·tan B / tan A+tan B)
⇒tan A ·tan B,
LHS=RHS
hence proved

Answer

Cot A + cot B = (tan A + tan B)/tan A tan B
So,
(tan A + tan B)/(cot A + cot B) = tan A × tan B

(I think it is tan A × tan B and not tan A + tan B)

Answer

Proof of the Trigonometric Identity: tanA + tanB / cotA + cotB = tanA + tanB
The given trigonometric identity can be easily proved by manipulating the expressions using the definitions of trigonometric ratios.

Step 1: Convert tan and cot to sin and cos
Using the definitions of tan and cot, we can rewrite the expression as:
(tanA + tanB) / (cotA + cotB) = (sinA/cosA + sinB/cosB) / (cosA/sinA + cosB/sinB)

Step 2: Combine the fractions
Next, we simplify the expression by combining the fractions:
(sinA/cosA + sinB/cosB) / (cosA/sinA + cosB/sinB) = ((sinAcosB + sinBcosA) / (cosAcosB)) / ((cosAsinB + sinAcosB) / (sinAcosB))
= (sin(A+B) / cos(A+B)) / (cos(A+B) / sin(A+B))

Step 3: Simplify the expression
After simplifying, we get:
(sin(A+B) / cos(A+B)) / (cos(A+B) / sin(A+B)) = (sin(A+B) * sin(A+B)) / (cos(A+B) * cos(A+B))
= (sin^2(A+B)) / (cos^2(A+B))

Step 4: Use trigonometric identity
Finally, we use the trigonometric identity sin^2(x) = 1 - cos^2(x) to simplify the expression further:
(sin^2(A+B)) / (cos^2(A+B)) = (1 - cos^2(A+B)) / cos^2(A+B)
= 1/cos^2(A+B) - cos^2(A+B)/cos^2(A+B)
= sec^2(A+B) - 1
Therefore, tanA + tanB / cotA + cotB is equal to sec^2(A+B) - 1, which proves the given trigonometric identity.

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