Eliminating θ from Trigonometric Equations | Class 10 Maths

Quick answer: To eliminate $\theta$ from two trigonometric equations, square them and then add (for sin and cos) or subtract (for sec and tan, or cosec and cot) so that the cross-terms cancel and an identity such as $\sin^2\theta + \cos^2\theta = 1$ or $\sec^2\theta - \tan^2\theta = 1$ removes $\theta$. For example, if $x = a\cos\theta + b\sin\theta$ and $y = a\sin\theta - b\cos\theta$, then $x^2 + y^2 = a^2 + b^2$.

What does "eliminate θ" mean?

You are given two equations that both contain $\theta$. To eliminate $\theta$ means to combine them into one equation that does not contain $\theta$ at all. You do not need to find the value of $\theta$.

The three identities you need

Problem 1: square and add

If $x = a\cos\theta + b\sin\theta$ and $y = a\sin\theta - b\cos\theta$, prove that $x^2 + y^2 = a^2 + b^2$.

Step 1: Square the first equation.

$x^2 = a^2\cos^2\theta + b^2\sin^2\theta + 2ab\sin\theta\cos\theta$

Step 2: Square the second equation.

$y^2 = a^2\sin^2\theta + b^2\cos^2\theta - 2ab\sin\theta\cos\theta$

Step 3: Add. The terms $+2ab\sin\theta\cos\theta$ and $-2ab\sin\theta\cos\theta$ cancel:

$x^2 + y^2 = a^2(\cos^2\theta + \sin^2\theta) + b^2(\sin^2\theta + \cos^2\theta)$

Step 4: Use the identity $\sin^2\theta + \cos^2\theta = 1$:

$x^2 + y^2 = a^2 + b^2$

Hence proved.

Problem 2: square and subtract

If $x = a\sec\theta + b\tan\theta$ and $y = a\tan\theta + b\sec\theta$, prove that $x^2 - y^2 = a^2 - b^2$.

Step 1: Square both equations.

$x^2 = a^2\sec^2\theta + b^2\tan^2\theta + 2ab\sec\theta\tan\theta$

$y^2 = a^2\tan^2\theta + b^2\sec^2\theta + 2ab\sec\theta\tan\theta$

Step 2: Subtract. The cross-terms are the same, so they cancel:

$x^2 - y^2 = a^2(\sec^2\theta - \tan^2\theta) - b^2(\sec^2\theta - \tan^2\theta)$

Step 3: Use $\sec^2\theta - \tan^2\theta = 1$:

$x^2 - y^2 = a^2 - b^2$

Hence proved.

Problem 3: prove m² − n² = 4√(mn)

If $\tan\theta + \sin\theta = m$ and $\tan\theta - \sin\theta = n$, prove that $m^2 - n^2 = 4\sqrt{mn}$.

Step 1: Left-hand side. Use $m^2 - n^2 = (m + n)(m - n)$.

• $m + n = 2\tan\theta$
• $m - n = 2\sin\theta$

$\text{LHS} = (2\tan\theta)(2\sin\theta) = 4\tan\theta\sin\theta$

Step 2: Find mn.

$mn = (\tan\theta + \sin\theta)(\tan\theta - \sin\theta) = \tan^2\theta - \sin^2\theta$

Write $\tan^2\theta$ as $\frac{\sin^2\theta}{\cos^2\theta}$ and take $\sin^2\theta$ common:

$mn = \sin^2\theta\left(\frac{1}{\cos^2\theta} - 1\right) = \sin^2\theta(\sec^2\theta - 1)$

Since $\sec^2\theta - 1 = \tan^2\theta$:

$mn = \sin^2\theta\tan^2\theta$

Step 3: Right-hand side. For an acute angle $\theta$, $\sin\theta$ and $\tan\theta$ are positive, so:

$\text{RHS} = 4\sqrt{\sin^2\theta\tan^2\theta} = 4\sin\theta\tan\theta$

Step 4: Compare. LHS = RHS = $4\sin\theta\tan\theta$, so $m^2 - n^2 = 4\sqrt{mn}$.

Hence proved.

Useful result: $\tan^2\theta - \sin^2\theta = \tan^2\theta\sin^2\theta$. Spotting this saves several lines of working.

Summary table

Exam tips

• Look at the cross-terms after squaring: opposite signs → add, same signs → subtract.
• In "prove that" questions, simplify the LHS and RHS separately and show both reach the same expression.
• State the identity you use at each step, for example "since sec²θ − tan²θ = 1".
• Mention that $\theta$ is acute when you take a square root, so the signs are positive.

Common mistakes

• Trying to find $\theta$ using a calculator or inverse functions. You only need to remove it.
• Forgetting the $2ab$ cross-term when squaring a sum such as $(a\cos\theta + b\sin\theta)^2$.
• Writing $\sec^2\theta + \tan^2\theta = 1$. The correct identity is $\sec^2\theta - \tan^2\theta = 1$.
• Writing $1 - \sec^2\theta = \tan^2\theta$; it equals $-\tan^2\theta$.

Frequently Asked Questions

What does it mean to eliminate θ?

It means combining two equations that contain θ into a single equation without θ, usually by squaring and using a trigonometric identity.

When should I add and when should I subtract after squaring?

For sin and cos, add, because sin²θ + cos²θ = 1. For sec and tan, or cosec and cot, subtract, because sec²θ − tan²θ = 1 and cosec²θ − cot²θ = 1.

How do you prove x² + y² = a² + b² when x = a cos θ + b sin θ?

Square both equations and add them. The 2ab sin θ cos θ terms cancel, and sin²θ + cos²θ = 1 leaves a² + b².

How do you prove m² − n² = 4√(mn)?

Show that m² − n² = (m + n)(m − n) = 4 tan θ sin θ, and that mn = tan²θ − sin²θ = sin²θ tan²θ, so 4√(mn) is also 4 sin θ tan θ.

Which trigonometric identities are used for elimination?

The three Pythagorean identities: sin²θ + cos²θ = 1, sec²θ − tan²θ = 1 and cosec²θ − cot²θ = 1.