Question

In a triangle ABC, tanA2=56
and tanC2=25
, then

a, b, c are in A.P.

cos A, cos B, cos C are in A.P.

sin A, sin B, sin C are in A.P.

Both (a) and (c)

Answer 1

Since A + B + C = 180°, we can use the identity tan(A + B + C) = 0.

We have tan(A + B + C) = (tanA + tanB + tanC - tanA*tanB*tanC)/(1 - tanA*tanB - tanB*tanC - tanA*tanC)

Plugging in the given values, we have:

0 = (tanA + tanB + tanC - tanA*tanB*tanC)/(1 - tanA*tanB - tanB*tanC - tanA*tanC)

Multiplying both sides by (1 - tanA*tanB - tanB*tanC - tanA*tanC), we get:

0 = tanA + tanB + tanC - tanA*tanB*tanC

Simplifying, we have:

tanA*tanB*tanC = tanA + tanB + tanC

Since a, b, c are in an arithmetic progression, we have:

b = (a + c)/2

Squaring both sides, we get:

b^2 = (a^2 + 2ac + c^2)/4

Replacing a with tanA^2 and c with tanC^2, we get:

b^2 = (tanA^2 + 2*tanA^2*tanC^2 + tanC^2)/4

Using the given values tanA^2 = 56 and tanC^2 = 25, we can simplify the equation to:

b^2 = (56 + 2*56*25 + 25)/4
b^2 = (56 + 2800 + 25)/4
b^2 = 2881/4
b = sqrt(2881)/2

Since b is the cosine of angle B, we have:

cosB = sqrt(2881)/2

Therefore, cos A, cos B, cos C are in an arithmetic progression.

Similarly, using the equation tanA*tanB*tanC = tanA + tanB + tanC, we can write:

sinA*sinB*sinC = sinA + sinB + sinC

Since a, b, c are in an arithmetic progression, we have:

b = (a + c)/2

Replacing a with sinA and c with sinC, we get:

b = (sinA + sinC)/2

Squaring both sides, we get:

b^2 = (sinA^2 + 2*sinA*sinC + sinC^2)/4

Using the given values sinA^2 = 1 - cosA^2 and sinC^2 = 1 - cosC^2, we can simplify the equation to:

b^2 = (1 - cosA^2 + 2*sinA*sinC + 1 - cosC^2)/4
b^2 = (2 - cosA^2 - cosC^2 + 2*sinA*sinC)/4
b^2 = (2 - cosA^2 - cosC^2 + 2*sinA*sinC)/4
b^2 = (2 - cosA^2 - (1 - cosA^2) + 2*sinA*(sqrt(1 - cosA^2))/4
b^2 = (2 + 2*sinA*(sqrt(1 - cosA^2))/4
b^2 = (1/2) + sinA*(sqrt(1 - cosA^2))

Similarly, using the given values tanA^2 = 56 and tanC^2 = 25, we can rewrite the equation tanA*tanB*tanC = tanA + tanB + tanC as:

(sqrt(1 - cosA^2))*(sqrt(1 - cosB^2))*(sqrt(1 - cosC^2)) = sqrt(1 - cosA^2) + sqrt(1 - cosB^2) + sqrt(1 - cosC^2)

Simplifying, we have:

sqrt(1 - cosA^2)*sqrt(1 - cosB^2)*sqrt(1 - cosC^2) = sqrt(1 - cosA^2) + sqrt(1 - cosB^2) + sqrt(1 - cosC^2)

Multiplying both sides by sqrt(1 - cosA^2)*sqrt(1 - cosB^2)*sqrt(1 - cosC^2), we get:

1 - cosA^2 - cosB^2 - cosC^2 + cosA^2*cosB^2 + cosB^2*cosC^2 + cosA^2*cosC^2 - cosA^2*cosB - cosB^2*cosC - cosA*cosC^2 + cosA*cosC + cosA*cosB - cosA^2 + cosB^2 + cosC^2 - cosA + cosB - cosC = 1 - cosA^2 + 2*sqrt(1 - cosA^2) + 1 - cosB^2 + 2*sqrt(1 - cosB^2) + 1 - cosC^2 + 2*sqrt(1 - cosC^2)

Simplifying and rearranging, we get:

2*sqrt(1 - cosA^2) + 2*sqrt(1 - cosB^2) + 2*sqrt(1 - cosC^2) = 2*cosA*cosB*cosC - cosA^2*cosB^2 - cosB^2*cosC^2 - cosA^2*cosC^2 + cosA*cosB + cosB*cosC + cosA*cosC - cosA - cosB - cosC

Dividing both sides by 2, we get:

sqrt(1 - cosA^2) + sqrt(1 - cosB^2) + sqrt(1 - cosC^2) = cosA*cosB*cosC - (1/2)*(cosA^2*cosB^2 + cosB^2*cosC^2 + cosA^2*cosC^2) + (1/2)*(cosA*cosB + cosB*cosC + cosA*cosC - cosA - cosB - cosC)

Since cos A, cos B, cos C are in an arithmetic progression, we have:

cosB = (cosA + cosC)/2

Squaring both sides, we get:

cosB^2 = (cosA^2 + 2*cosA*cosC + cosC^2)/4

Replacing a with cosA^2 and c with cosC^2, we get:

cosB^2 = (cosA^2 + 2*cosA*cosC + cosC^2)/4

Using the given values cosA^2 = 1 - sinA^2 and cosC^2 = 1 - sinC^2, we can simplify the equation to:

cosB^2 = (1 - sinA^2 + 2*sinA*sinC + 1 - sinC^2)/4
cosB^2 = (2 - sinA^2 - sinC^2 + 2*sinA*sinC)/4
cosB^2 = (2 - (1 - cosA^2) - (1 - cosC^2) + 2*sinA*sinC)/4
cosB^2 = (1/2) + sinA*sinC

Therefore, sin A, sin B, sin C are in an arithmetic progression.

Both (a) and (c) are true.