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Prove that (Tan A)/(sec A-1)-(1+Tan A)/(sec A+1)=2cosec theta

Prove that Tan⁡Asec⁡A−1−1+Tan⁡Asec⁡A+1=2cosec⁡θ \frac{\operatorname{Tan} A}{\sec A-1}-\frac{1+\operatorname{Tan} A}{\sec A+1}=2 \operatorname{cosec} \theta

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Q. Prove that Tan⁡Asec⁡A−1−1+Tan⁡Asec⁡A+1=2cosec⁡θ \frac{\operatorname{Tan} A}{\sec A-1}-\frac{1+\operatorname{Tan} A}{\sec A+1}=2 \operatorname{cosec} \theta
  1. Express tan and sec: Let's start by expressing tan⁡A\tan A and sec⁡A\sec A in terms of sin⁡A\sin A and cos⁡A\cos A, since tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A} and sec⁡A=1cos⁡A\sec A = \frac{1}{\cos A}.
  2. Rewrite using sin and cos: Rewrite the given expression using sin AA and cos AA: tan⁡Asec⁡A−1−1+tan⁡Asec⁡A+1=sin⁡A/cos⁡A1/cos⁡A−1−1+sin⁡A/cos⁡A1/cos⁡A+1\frac{\tan A}{\sec A-1}-\frac{1+\tan A}{\sec A+1} = \frac{\sin A / \cos A}{1/\cos A - 1} - \frac{1 + \sin A / \cos A}{1/\cos A + 1}.
  3. Simplify fractions: Simplify the denominators of both fractions by multiplying by cos⁡A\cos A:(sin⁡Acos⁡A)/(1cos⁡A−1)\left(\frac{\sin A}{\cos A}\right) / \left(\frac{1}{\cos A} - 1\right) becomes (sin⁡Acos⁡A)∗(cos⁡A1−cos⁡A)\left(\frac{\sin A}{\cos A}\right) * \left(\frac{\cos A}{1 - \cos A}\right), and (1+sin⁡Acos⁡A)/(1cos⁡A+1)\left(1 + \frac{\sin A}{\cos A}\right) / \left(\frac{1}{\cos A} + 1\right) becomes (1+sin⁡Acos⁡A)∗(cos⁡A1+cos⁡A)\left(1 + \frac{\sin A}{\cos A}\right) * \left(\frac{\cos A}{1 + \cos A}\right).
  4. Combine and subtract: Now, simplify the expressions:\newline(sin⁡Acos⁡A)⋅(cos⁡A1−cos⁡A)=sin⁡A1−cos⁡A(\frac{\sin A}{\cos A}) \cdot (\frac{\cos A}{1 - \cos A}) = \frac{\sin A}{1 - \cos A},\newlineand (1+sin⁡Acos⁡A)⋅(cos⁡A1+cos⁡A)=cos⁡A+sin⁡A1+cos⁡A(1 + \frac{\sin A}{\cos A}) \cdot (\frac{\cos A}{1 + \cos A}) = \frac{\cos A + \sin A}{1 + \cos A}.
  5. Find common denominator: Subtract the second expression from the first:\newline(sin⁡A1−cos⁡A)−(cos⁡A+sin⁡A1+cos⁡A)(\frac{\sin A}{1 - \cos A}) - (\frac{\cos A + \sin A}{1 + \cos A}).
  6. Expand numerator: Find a common denominator to combine the fractions: [(sin⁡A)(1+cos⁡A)−(cos⁡A+sin⁡A)(1−cos⁡A)]/((1−cos⁡A)(1+cos⁡A))[(\sin A)(1 + \cos A) - (\cos A + \sin A)(1 - \cos A)] / ((1 - \cos A)(1 + \cos A)).
  7. Simplify numerator: Expand the numerator: (sin⁡A+sin⁡Acos⁡A−cos⁡A−sin⁡A+sin⁡Acos⁡A−sin⁡A2)/((1−cos⁡A)(1+cos⁡A))(\sin A + \sin A \cos A - \cos A - \sin A + \sin A \cos A - \sin A^2) / ((1 - \cos A)(1 + \cos A)).
  8. Replace with identity: Simplify the numerator by combining like terms: (sin⁡Acos⁡A+sin⁡Acos⁡A−cos⁡A−sin⁡A2)/((1−cos⁡A)(1+cos⁡A))(\sin A \cos A + \sin A \cos A - \cos A - \sin A^2) / ((1 - \cos A)(1 + \cos A)).
  9. Simplify denominator: Notice that sin⁡Acos⁡A+sin⁡Acos⁡A\sin A \cos A + \sin A \cos A is 2sin⁡Acos⁡A2\sin A \cos A, and sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1: (2sin⁡Acos⁡A−cos⁡A−(1−cos⁡2A))/((1−cos⁡A)(1+cos⁡A)).(2\sin A \cos A - \cos A - (1 - \cos^2 A)) / ((1 - \cos A)(1 + \cos A)).
  10. Divide by sin⁡2A\sin^2 A: Replace 1−cos⁡2A1 - \cos^2 A with sin⁡2A\sin^2 A, which is an identity:\newline(2sin⁡Acos⁡A−cos⁡A−sin⁡2A)/((1−cos⁡A)(1+cos⁡A))(2\sin A \cos A - \cos A - \sin^2 A) / ((1 - \cos A)(1 + \cos A)).
  11. Final simplification: Now, simplify the denominator using the difference of squares identity: \newlineegin{equation}\newline\frac{22\sin A \cos A - \cos A - \sin^22 A}{11 - \cos^22 A}.\newline\end{equation}
  12. Final simplification: Now, simplify the denominator using the difference of squares identity: \newline(2sin⁡Acos⁡A−cos⁡A−sin⁡2A)/(1−cos⁡2A)(2\sin A \cos A - \cos A - \sin^2 A) / (1 - \cos^2 A).Replace 1−cos⁡2A1 - \cos^2 A with sin⁡2A\sin^2 A in the denominator:\newline(2sin⁡Acos⁡A−cos⁡A−sin⁡2A)/sin⁡2A(2\sin A \cos A - \cos A - \sin^2 A) / \sin^2 A.
  13. Final simplification: Now, simplify the denominator using the difference of squares identity: \newline(2sin⁡Acos⁡A−cos⁡A−sin⁡2A)/(1−cos⁡2A)(2\sin A \cos A - \cos A - \sin^2 A) / (1 - \cos^2 A).Replace 1−cos⁡2A1 - \cos^2 A with sin⁡2A\sin^2 A in the denominator: \newline(2sin⁡Acos⁡A−cos⁡A−sin⁡2A)/sin⁡2A(2\sin A \cos A - \cos A - \sin^2 A) / \sin^2 A.Divide each term in the numerator by sin⁡2A\sin^2 A: \newline(2cos⁡A/sin⁡A)−(1/sin⁡A)−(sin⁡2A/sin⁡2A)(2\cos A/\sin A) - (1/\sin A) - (\sin^2 A/\sin^2 A).
  14. Final simplification: Now, simplify the denominator using the difference of squares identity: \newline(2sin⁡Acos⁡A−cos⁡A−sin⁡2A)/(1−cos⁡2A)(2\sin A \cos A - \cos A - \sin^2 A) / (1 - \cos^2 A).Replace 1−cos⁡2A1 - \cos^2 A with sin⁡2A\sin^2 A in the denominator: \newline(2sin⁡Acos⁡A−cos⁡A−sin⁡2A)/sin⁡2A(2\sin A \cos A - \cos A - \sin^2 A) / \sin^2 A.Divide each term in the numerator by sin⁡2A\sin^2 A: \newline(2cos⁡A/sin⁡A)−(1/sin⁡A)−(sin⁡2A/sin⁡2A)(2\cos A/\sin A) - (1/\sin A) - (\sin^2 A/\sin^2 A).Simplify the terms: \newline(2cot⁡A)−(csc⁡A)−1(2\cot A) - (\csc A) - 1.
  15. Final simplification: Now, simplify the denominator using the difference of squares identity:\newline(2sin⁡Acos⁡A−cos⁡A−sin⁡2A)/(1−cos⁡2A)(2\sin A \cos A - \cos A - \sin^2 A) / (1 - \cos^2 A).Replace 1−cos⁡2A1 - \cos^2 A with sin⁡2A\sin^2 A in the denominator:\newline(2sin⁡Acos⁡A−cos⁡A−sin⁡2A)/sin⁡2A(2\sin A \cos A - \cos A - \sin^2 A) / \sin^2 A.Divide each term in the numerator by sin⁡2A\sin^2 A:\newline(2cos⁡A/sin⁡A)−(1/sin⁡A)−(sin⁡2A/sin⁡2A)(2\cos A/\sin A) - (1/\sin A) - (\sin^2 A/\sin^2 A).Simplify the terms:\newline(2cot⁡A)−(csc⁡A)−1(2\cot A) - (\csc A) - 1.We have made a mistake in the previous step. The term (sin⁡2A/sin⁡2A)(\sin^2 A/\sin^2 A) should simplify to 11, not −1-1. Let's correct this:\newline(2cot⁡A)−(csc⁡A)−1(2\cot A) - (\csc A) - 1 becomes 1−cos⁡2A1 - \cos^2 A11.

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