verify the following identity:
cos(A-B)/cosAsinB=tanA+cotB
To verify the given identity cos(A-B) / (cosA * sinB) = tanA + cotB, we can manipulate the expression on both sides of the equation to show that they are equal.
Starting with the left side of the equation: cos(A-B) / (cosA * sinB)
We can use the cosine difference formula to simplify cos(A-B):
cos(A-B) = cosA*cosB + sinA*sinB
Substituting back into the expression on the left side:
(cosA*cosB + sinA*sinB) / (cosA * sinB)
We can then factor out cosA from the numerator:
cosA(cosB / (cosA * sinB) + (sinA*sinB) / (cosA * sinB))
Now, rearranging the terms in the numerator:
cosA((cosB + sinA*sinB) / (cosA * sinB))
Simplifying the expression inside the brackets:
cosA((cosB + sinA*sinB) / (sinA * sinB))
Next, canceling the sinB terms in the numerator:
cosA((cosB + sinA) / sinA)
Expanding the numerator:
cosA * cosB / sinA + cosA * sinA / sinA
Simplifying the second term:
cosA * cosB / sinA + cosA
Now, expressing cosA as 1/ secA:
(1 / secA) * cosB / sinA + cosA
Using the definition of tangent as sinA / cosA:
(1 / secA) * cosB / sinA + tanA
Multiplying the numerator by cosA and the denominator by secA to simplify:
cosB / sinA + tanA
Finally, using the reciprocal identity of cotangent:
cotB + tanA
Therefore, we can conclude that the left side is equal to the right side of the original equation, and the identity is verified.