Question

tanA+tan(90+A)+tan(180-A)+ tan(90-A)​

Answer 1

tanA+tan(90+A)+tan(180-A)+ tan(90-A)​ =0

• We have to find the value of tanA + tan(90+A) + tan(180+A) + tan(90-A)

• tan(90+A) = sin(90+A)/cos(90+A)

sin(A+B) = sinA.cosB + cosA.sinB

cos(A+B) = cosA.cosB - sinA.sinB

• Now, tan(90+A) = sin(90+A)/cos(90+A)

= (sin90.cosA + cos90.sinA)/(cos90.cosA - sin90.sinA)

= cosA / (-sinA) = -cotA

• Similarly , tan(90-A) = sin(90-A)/cos(90-A)

= (sin90.cosA - cos90.sinA)/(cos90.cosA + sin90.sinA)

= cosA / (sinA) = cotA

• And tan(180-A) = -tanA

• Now substituting the values we obtained in the equation, we get

tanA + tan(90+A) + tan(180+A) + tan(90-A)

= tanA + (-cotA) + (-tanA) + (cotA) = 0

Answer 2

Answer:

answer is zero

Step-by-step explanation:

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