Maths Q20 – AP-EAMCET 2026

Home Maths Trigonometric FunctionsQuestion #479

\cos x = \tan y

\cos y = \tan z

and

\cos z = \tan x

, then a value of

\sin x

Clue 1 of 3: Cyclic Substitution

Observe the cyclic nature of the given equations: $\cos x = \tan y$ , $\cos y = \tan z$ , $\cos z = \tan x$ . This suggests a substitution strategy.

🧠 Full Solution Path

Step 1: Transforming the given equations using identities✦ Active

The given equations are:

\cos x = \tan y (1)

\cos y = \tan z (2)

\cos z = \tan x (3)

Square each equation:

\cos^{2} x = \tan^{2} y (1^{'})

\cos^{2} y = \tan^{2} z (2^{'})

\cos^{2} z = \tan^{2} x (3^{'})

Use the identity $\tan^{2} θ = \sec^{2} θ - 1 = \frac{1}{\cos^{2} θ} - 1$ . Substitute this into $(1^{'}), (2^{'}), (3^{'})$ :

\cos^{2} x = \frac{1}{\cos^{2} y} - 1 ⟹ \cos^{2} x \cos^{2} y = 1 - \cos^{2} y = \sin^{2} y (A)

\cos^{2} y = \frac{1}{\cos^{2} z} - 1 ⟹ \cos^{2} y \cos^{2} z = 1 - \cos^{2} z = \sin^{2} z (B)

\cos^{2} z = \frac{1}{\cos^{2} x} - 1 ⟹ \cos^{2} z \cos^{2} x = 1 - \cos^{2} x = \sin^{2} x (C)

💡 Teacher's Secret Hint

Note that since cosine values are between -1 and 1, the tangent values must also be between -1 and 1. This implies that $x, y, z$ must be in the range $[- π / 4, π / 4]$ . If we assume $x, y, z$ are acute angles, then $x, y, z \in (0, π / 4]$ .

Step 2: Setting up a system of equations for

\cos^{2} x, \cos^{2} y, \cos^{2} z

○ Expand

From equations (A), (B), (C), we can write:

\cos^{2} x = \frac{1}{1 + \cos^{2} y}

\cos^{2} y = \frac{1}{1 + \cos^{2} z}

\cos^{2} z = \frac{1}{1 + \cos^{2} x}

Let $A = \cos^{2} x$ , $B = \cos^{2} y$ , $C = \cos^{2} z$ . Then we have the system:

A = \frac{1}{1 + B} (i)

B = \frac{1}{1 + C} (i i)

C = \frac{1}{1 + A} (i i i)

Step 3: Solving for

\cos^{2} x

○ Expand

Substitute $(i i)$ into $(i)$ :

A = \frac{1}{1 + \frac{1}{1 + C}} = \frac{1}{\frac{1 + C + 1}{1 + C}} = \frac{1 + C}{2 + C}

Now substitute $(i i i)$ into this expression for $A$ :

A = \frac{1 + \frac{1}{1 + A}}{2 + \frac{1}{1 + A}} = \frac{\frac{1 + A + 1}{1 + A}}{\frac{2 (1 + A) + 1}{1 + A}} = \frac{2 + A}{2 A + 3}

Cross-multiply and simplify:

A (2 A + 3) = 2 + A

2 A^{2} + 3 A = 2 + A

2 A^{2} + 2 A - 2 = 0

A^{2} + A - 1 = 0

This is a quadratic equation for $A = \cos^{2} x$ . Using the quadratic formula:

A = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (- 1)}}{2 (1)} = \frac{- 1 \pm \sqrt{1 + 4}}{2} = \frac{- 1 \pm \sqrt{5}}{2}

Since $A = \cos^{2} x$ must be positive, we take the positive root:

\cos^{2} x = \frac{\sqrt{5} - 1}{2}

Step 4: Calculating

\sin x

○ Expand

We need to find $\sin x$ . We know $\sin^{2} x = 1 - \cos^{2} x$ .

\sin^{2} x = 1 - (\frac{\sqrt{5} - 1}{2}) = \frac{2 - (\sqrt{5} - 1)}{2} = \frac{2 - \sqrt{5} + 1}{2} = \frac{3 - \sqrt{5}}{2}

Since $x \in (0, π / 4]$ , $\sin x$ must be positive.

\sin x = \sqrt{\frac{3 - \sqrt{5}}{2}}

We know that $2 \sin 18^{\circ} = 2 (\frac{\sqrt{5} - 1}{4}) = \frac{\sqrt{5} - 1}{2}$ . Let's check if $\sqrt{\frac{3 - \sqrt{5}}{2}}$ is equal to $\frac{\sqrt{5} - 1}{2}$ .

Square both sides:

{(\sqrt{\frac{3 - \sqrt{5}}{2}})}^{2} = \frac{3 - \sqrt{5}}{2}

{(\frac{\sqrt{5} - 1}{2})}^{2} = \frac{(\sqrt{5})^{2} - 2 \sqrt{5} + 1^{2}}{4} = \frac{5 - 2 \sqrt{5} + 1}{4} = \frac{6 - 2 \sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}

Since both sides are equal, we have:

\sin x = \frac{\sqrt{5} - 1}{2}

This value is equal to $2 \sin 18^{\circ}$ .

💡 Teacher's Secret Hint

The value $\frac{\sqrt{5} - 1}{2}$ is a common trigonometric constant, often appearing in relation to angles like $18^{\circ}$ or $36^{\circ}$ . Specifically, it is equal to $2 \sin 18^{\circ}$ or $\cos 36^{\circ}$ .

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Maths Question 20 – AP-EAMCET 2026

Clue 1 of 3: Cyclic Substitution

🧠 Full Solution Path