Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Prove the identities with all angles acute
i. 
(1+sec A)/(sec A)=(sin^(2)A)/(1-cos A)

Prove the identities with all angles acute\newlinei. 1+secAsecA=sin2A1cosA \frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}

Full solution

Q. Prove the identities with all angles acute\newlinei. 1+secAsecA=sin2A1cosA \frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}
  1. Rewrite using secant definition: Rewrite the left side of the equation using the definition of secant: secA=1cosA\sec A = \frac{1}{\cos A}. So, (1+secA)/(secA)(1+\sec A)/(\sec A) becomes (1+1cosA)/(1cosA)(1 + \frac{1}{\cos A})/(\frac{1}{\cos A}).
  2. Combine terms in numerator: Combine the terms in the numerator: 1+1cosA=cosA+1cosA.1 + \frac{1}{\cos A} = \frac{\cos A + 1}{\cos A}.
  3. Divide numerator by denominator: Now, divide the combined numerator by the denominator: (cosA+1cosA)/(1cosA)\left(\frac{\cos A + 1}{\cos A}\right) / \left(\frac{1}{\cos A}\right).
  4. Simplify division by reciprocal: Simplify the division by multiplying by the reciprocal of the denominator: (cosA+1)/cosA×cosA/1(\cos A + 1)/\cos A \times \cos A/1.
  5. Cancel out cosA\cos A terms: The cosA\cos A terms cancel out, leaving us with cosA+1\cos A + 1.
  6. Use Pythagorean identity: Now, let's look at the right side of the equation: sin2A1cosA\frac{\sin^{2}A}{1-\cos A}.
  7. Substitute identity into right side: Use the Pythagorean identity sin2A=1cos2A\sin^2 A = 1 - \cos^2 A.
  8. Factor numerator as difference of squares: Substitute the identity into the right side of the equation: (1cos2A)/(1cosA)(1 - \cos^2 A)/(1 - \cos A).
  9. Cancel out (1cosA)(1 - \cos A) terms: Factor the numerator as a difference of squares: (1cosA)(1+cosA)1cosA\frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A}.
  10. Final expression on both sides: The (1cosA)(1 - \cos A) terms cancel out, leaving us with 1+cosA1 + \cos A.
  11. Final expression on both sides: The (1cosA)(1 - \cos A) terms cancel out, leaving us with 1+cosA1 + \cos A.We now have the same expression on both sides of the equation: cosA+1\cos A + 1.

More problems from Sin, cos, and tan of special angles