Prove (cos A * Cot A) / (1 - Sin A) = 1 + Cosec A

by Rajiv Sharma 50 views

Hey guys! Let's dive into this awesome trigonometric proof. We're going to show that cos⁡A⋅cot⁡A1−sin⁡A\frac{\cos A \cdot \cot A}{1-\sin A} is indeed equal to 1+cosec⁡A1+\operatorname{cosec} A. Grab your pencils, and let's get started!

Breaking Down the Problem

Before we jump into the nitty-gritty, let's understand what we're dealing with. Our mission is to prove the trigonometric identity: cos⁡A⋅cot⁡A1−sin⁡A=1+cosec⁡A\frac{\cos A \cdot \cot A}{1-\sin A}=1+\operatorname{cosec} A. This means we'll start with the left-hand side (LHS) of the equation and manipulate it using trigonometric identities until we arrive at the right-hand side (RHS). It's like a puzzle, and the identities are our tools!

Let's quickly recap some essential trigonometric functions and identities that will help us in this proof:

  • Cosine (cos A): The ratio of the adjacent side to the hypotenuse in a right-angled triangle.
  • Cotangent (cot A): The ratio of the adjacent side to the opposite side, and it's also the reciprocal of the tangent function. Mathematically, cot⁡A=cos⁡Asin⁡A\cot A = \frac{\cos A}{\sin A}.
  • Cosecant (cosec A): The reciprocal of the sine function. Mathematically, cosec⁡A=1sin⁡A\operatorname{cosec} A = \frac{1}{\sin A}.
  • Pythagorean Identity: The most famous one! sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1. We'll probably use this.

Now that we've got our toolkit ready, let's roll up our sleeves and tackle the left-hand side of the equation.

Step-by-Step Proof

1. Start with the Left-Hand Side (LHS)

The first step is always to begin with the expression we want to simplify, which is the left-hand side (LHS) of our equation:

cos⁡A⋅cot⁡A1−sin⁡A\qquad \frac{\cos A \cdot \cot A}{1-\sin A}

2. Substitute cot⁡A\cot A

Remember that cot⁡A\cot A can be written as cos⁡Asin⁡A\frac{\cos A}{\sin A}. Let's substitute this into our LHS expression:

cos⁡A⋅cos⁡Asin⁡A1−sin⁡A\qquad \frac{\cos A \cdot \frac{\cos A}{\sin A}}{1-\sin A}

This substitution is crucial because it expresses everything in terms of sines and cosines, making it easier to manipulate.

3. Simplify the Numerator

Now, let's simplify the numerator by multiplying the cosine terms:

cos⁡2Asin⁡A1−sin⁡A\qquad \frac{\frac{\cos^2 A}{\sin A}}{1-\sin A}

This simplifies our expression, bringing us closer to our goal.

4. Multiply by sin⁡Asin⁡A\frac{\sin A}{\sin A} (Clever Trick!)

To get rid of the fraction within a fraction, we're going to use a sneaky trick. We'll multiply the numerator and the denominator by sin⁡A\sin A. This is equivalent to multiplying by 1, so it doesn't change the value of the expression, but it sure cleans things up:

cos⁡2Asin⁡A1−sin⁡A⋅sin⁡Asin⁡A=cos⁡2Asin⁡A(1−sin⁡A)\qquad \frac{\frac{\cos^2 A}{\sin A}}{1-\sin A} \cdot \frac{\sin A}{\sin A} = \frac{\cos^2 A}{\sin A(1-\sin A)}

See how we've eliminated the fraction in the numerator? Nice!

5. Use the Pythagorean Identity

Remember the Pythagorean identity: sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1? We can rearrange this to solve for cos⁡2A\cos^2 A:

cos⁡2A=1−sin⁡2A\qquad \cos^2 A = 1 - \sin^2 A

Let's substitute this into our expression:

1−sin⁡2Asin⁡A(1−sin⁡A)\qquad \frac{1 - \sin^2 A}{\sin A(1-\sin A)}

This is a big step because it introduces a term that we can factor.

6. Factor the Numerator

Notice that the numerator, 1−sin⁡2A1 - \sin^2 A, is a difference of squares. We can factor it as follows:

1−sin⁡2A=(1−sin⁡A)(1+sin⁡A)\qquad 1 - \sin^2 A = (1 - \sin A)(1 + \sin A)

Now, substitute the factored form back into our expression:

(1−sin⁡A)(1+sin⁡A)sin⁡A(1−sin⁡A)\qquad \frac{(1 - \sin A)(1 + \sin A)}{\sin A(1-\sin A)}

Factoring is a powerful tool in simplifying trigonometric expressions.

7. Cancel Common Factors

Aha! We now have a common factor of (1−sin⁡A)(1 - \sin A) in both the numerator and the denominator. Let's cancel them out:

(1−sin⁡A)(1+sin⁡A)sin⁡A(1−sin⁡A)=1+sin⁡Asin⁡A\qquad \frac{\cancel{(1 - \sin A)}(1 + \sin A)}{\sin A\cancel{(1-\sin A)}} = \frac{1 + \sin A}{\sin A}

Canceling common factors simplifies the expression significantly.

8. Split the Fraction

We're almost there! Now, let's split the fraction into two separate fractions:

1+sin⁡Asin⁡A=1sin⁡A+sin⁡Asin⁡A\qquad \frac{1 + \sin A}{\sin A} = \frac{1}{\sin A} + \frac{\sin A}{\sin A}

Splitting fractions can often reveal simpler terms.

9. Simplify the Fractions

Now, let's simplify each fraction. We know that 1sin⁡A=cosec⁡A\frac{1}{\sin A} = \operatorname{cosec} A and sin⁡Asin⁡A=1\frac{\sin A}{\sin A} = 1:

1sin⁡A+sin⁡Asin⁡A=cosec⁡A+1\qquad \frac{1}{\sin A} + \frac{\sin A}{\sin A} = \operatorname{cosec} A + 1

10. Write the Final Result

Finally, we can rewrite the expression to match the right-hand side (RHS) of our original equation:

cosec⁡A+1=1+cosec⁡A\qquad \operatorname{cosec} A + 1 = 1 + \operatorname{cosec} A

Voilà! We have successfully transformed the left-hand side (LHS) into the right-hand side (RHS).

Conclusion

So guys, we've proven that cos⁡A⋅cot⁡A1−sin⁡A=1+cosec⁡A\frac{\cos A \cdot \cot A}{1-\sin A}=1+\operatorname{cosec} A. We started with the LHS, used trigonometric identities like cot⁡A=cos⁡Asin⁡A\cot A = \frac{\cos A}{\sin A} and the Pythagorean identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1, factored expressions, canceled common factors, and simplified fractions. Each step was crucial in getting us to the final result.

Trigonometric proofs can seem daunting at first, but breaking them down into smaller, manageable steps makes them much easier to handle. Remember, the key is to know your identities and to be patient. Keep practicing, and you'll become a pro at these in no time!

If you have any questions or want to try another proof, let me know! Keep up the awesome work!