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(cosec A-1)/(cosec A+1)=((cos A)/(1+sin A))^(2)

cosec⁡A−1cosec⁡A+1=(cos⁡A1+sin⁡A)2 \frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\left(\frac{\cos A}{1+\sin A}\right)^{2}

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Sin, cos, and tan of special anglesright chevron icon

Full solution

Q. cosec⁡A−1cosec⁡A+1=(cos⁡A1+sin⁡A)2 \frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\left(\frac{\cos A}{1+\sin A}\right)^{2}
  1. Identify Expression: Identify the expression on the left side of the equation.\newlineSimplify csc⁡A=1/sin⁡A\csc A = 1/\sin A.\newlineSubstitute into the left side: (csc⁡A−1)/(csc⁡A+1)=(1/sin⁡A−1)/(1/sin⁡A+1)(\csc A - 1)/(\csc A + 1) = (1/\sin A - 1)/(1/\sin A + 1).
  2. Simplify Left Side: Simplify the expression by finding a common denominator.\newlineRewrite as (1−sin⁡A)/(1+sin⁡A)(1 - \sin A)/(1 + \sin A).
  3. Find Common Denominator: Identify the expression on the right side of the equation.\newlineRewrite cos⁡A/(1+sin⁡A)\cos A/(1 + \sin A) and then square it: (cos⁡A/(1+sin⁡A))2(\cos A/(1 + \sin A))^2.
  4. Rewrite Right Side: Use the Pythagorean identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1 to express cos⁡2A\cos^2 A.\newlineSubstitute cos⁡2A=1−sin⁡2A\cos^2 A = 1 - \sin^2 A into the squared term: (cos⁡A/(1+sin⁡A))2=((1−sin⁡2A)/(1+sin⁡A))2(\cos A/(1 + \sin A))^2 = ((\sqrt{1 - \sin^2 A})/(1 + \sin A))^2.

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