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# Eliminate theta between a=cos(theta-alpha) and b=sin(theta-beta)?

We will be solving this question by using the concept of inverse trigonometric functions.

## Answer: We get cos^{-1}a - sin^{-1}b = beta - alpha after eliminating theta between a = cos(theta-alpha) and b = sin(theta-beta).

Let's look into the solution step by step

**Explanation:**

Given: a = cos(theta - alpha) and b = sin(theta - beta)

Let,

a = cos(theta - alpha) ----------------- (1)

b = sin(theta - beta) --------------- (2)

Now, apply cos^{-1 }on both the sides of (1)

⇒ cos^{-1}a = cos^{-1} cos(theta - alpha)

Since, we know that cos^{-1} cos(θ) = θ

⇒ cos^{-1}a = (theta - alpha) ------------------- (3)

Similarly, apply sin^{-1 }on both the sides of (2)

⇒ sin^{-1}b = sin^{-1 }sin(theta - beta)

Also, we know that sin^{-1} sin(θ) = θ

⇒ sin^{-1}b = (theta - beta) ------------------ (4)

Now, substracting (4) from (3) and we get,

cos^{-1}a - sin^{-1}b = (theta - alpha) - (theta - beta)

⇒ cos^{-1}a - sin^{-1}b = - alpha + beta

⇒ cos^{-1}a - sin^{-1}b = beta - alpha

### Hence, We get cos^{-1}a - sin^{-1}b = beta - alpha after eliminating theta between a = cos(theta-alpha) and b = sin(theta-beta)

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