Asked by Leah
Verify the identity & Justify the step.s
cot (theta-pi/2)=-tan theta
cot (theta-pi/2)=-tan theta
Answers
Answered by
Steve
cot(a-b) = (cota*cotb+1)/(cotb-cota)
cot(θ-π/2) = (cotθ*cotπ/2+1)/(cotπ/2-cotθ)
= (0+1)/(0-cotθ)
= -1/cotθ
= -tanθ
Also, note that cot(pi/2-x) = tan(x)
That's true for all the co-functions. That's why they are so named. They're the function of the complement.
sin(x) = cosine(pi/2-x)
same for tan and sec.
So, cot(x-pi/2) = -cot(pi/2 - x) = -tan(x)
cot(θ-π/2) = (cotθ*cotπ/2+1)/(cotπ/2-cotθ)
= (0+1)/(0-cotθ)
= -1/cotθ
= -tanθ
Also, note that cot(pi/2-x) = tan(x)
That's true for all the co-functions. That's why they are so named. They're the function of the complement.
sin(x) = cosine(pi/2-x)
same for tan and sec.
So, cot(x-pi/2) = -cot(pi/2 - x) = -tan(x)
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