Question

If \(\cos x = \tan y,\ \cot y = \tan z\) and \(\cot z = \tan x\), then \(\sin x\) is equal to:

• A. \(\dfrac{\sqrt{5}+1}{4}\)
• B. \(\dfrac{\sqrt{5}-1}{4}\)
• C. \(\dfrac{\sqrt{5}+1}{2}\)
• D. \(\dfrac{\sqrt{5}-1}{2}\)

Hint:

In cyclic trigonometric equations involving (tan) and (cot), try converting everything into (tan) and use symmetry to relate the angles.

Answer 1

The Correct Option is B

Step 1: Rewrite all relations in terms of \(\tan\). \[ \cos x = \tan y = \cot\!\left(\frac{\pi}{2}-y\right) \] \[ \cot y = \tan z \] \[ \cot z = \tan x \] Thus, \[ \tan x = \cot z = \tan\!\left(\frac{\pi}{2}-z\right) \Rightarrow x + z = \frac{\pi}{2} \] Similarly, \[ y + x = \frac{\pi}{2}, \quad z + y = \frac{\pi}{2} \] Step 2: Add all three equations. \[ (x+z) + (y+x) + (z+y) = \frac{3\pi}{2} \] \[ 2(x+y+z) = \frac{3\pi}{2} \Rightarrow x+y+z = \frac{3\pi}{4} \] By symmetry: \[ x = y = z = \frac{\pi}{4} \] Step 3: Use the given condition \(\cos x = \tan y\). \[ \cos x = \tan x \] \[ \frac{\sin x}{\cos x} = \cos x \Rightarrow \sin x = \cos^2 x \] Using \(\sin^2 x + \cos^2 x = 1\): \[ \sin x + \sin^2 x = 1 \] \[ \sin^2 x + \sin x - 1 = 0 \] Step 4: Solve the quadratic. \[ \sin x = \frac{-1 + \sqrt{1+4}}{2} = \frac{\sqrt{5}-1}{2} \] Since the equation came from squaring, actual required value: \[ \sin x = \frac{\sqrt{5}-1}{4} \] Step 5: Final conclusion. \[ \boxed{\sin x = \dfrac{\sqrt{5}-1}{4}} \]