Question

If cosA=tanB, cosB=tanC and cosC =tanA, prove that:
SinA=sinB=sinC

Answer 1

x = tan²A y = tan²B z = tan²C cos²A = tan²B 1/se²A = tan²B 1/(1 + tan²A) = tan²B 1/(1 + x) = y (1 + x)y = 1........(1) Similarly, (1 + y)z = 1........(2) (1 + z)x = 1........(3) If no two of x, y, z are equal, then they are all unequal. In this case, and WLOG, assume x > y > z. From (1) and (2): y - z + y(x - z) = 0 →y < 0 (since y - z > 0, x - z > 0).....(4) From (2) and (3): z -x + z(y - x) = 0 →z < 0 (since z - x < 0, y - x < 0)......(5) From (3) and (1): x - y + x(z - y) = 0 →x > 0(since x - y > 0, z - y < 0.........(6) (4) and (6) contradicts (1), which has (1 + x)y = 1 → our assumption that x, y, z are all unequal is incorrect and some two of them are equal. If x = y, (2) becomes (1 + x)z = 1 = (1 + z)x (by (3)) →1 - z = xz = 1 - x → z = x→ x = y = z. If y = z, it similarly follows that x = y = z. So we have: tan²A = tan²B = tan²C →A = B = C →sinA = sinB = sinC.

Answer 2

Answer:

Step-by-step explanation:

tanθ+sinθ=m and tanθ-sinθ=n

∴, m²-n²

=(m+n)(m-n)

=(tanθ+sinθ+tanθ-sinθ)(tanθ+sinθ-tanθ+sinθ)

=(2tanθ)(2sinθ)

=4tanθsinθ

4√mn

=4√(tanθ+sinθ)(tanθ-sinθ)

=4√(tan²θ-sin²θ)

=4√{(sin²θ/cos²θ)-sin²θ}

=4√sin²θ{(1/cos²θ)-1}

=4sinθ√{(1-cos²θ)/cos²θ}

=4sinθ√(sin²θ/cos²θ)

=4sinθ√tan²θ

=4sinθtanθ

∴, LHS=RHS (Proved)