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

Since, we know that cos^-1 cos(θ) = θ

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

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

⇒ sin^-1b = sin^-1 sin(theta - beta) 

Also, we know that sin^-1 sin(θ) = θ

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

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

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

⇒ cos^-1a - sin^-1b = - alpha + beta

⇒ cos^-1a - sin^-1b = beta - alpha

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