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Prove that: (cosec A- sec A) (cot A-tan A) = (cosec A + sec A) (sec A cosec A-2)​

User Cephalopod
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Answer: To prove the given identity:

(cosec A - sec A)(cot A - tan A) = (cosec A + sec A)(sec A cosec A - 2)

We will start by simplifying both sides of the equation.

Left-hand side (LHS):

(cosec A - sec A)(cot A - tan A)

Using trigonometric identities, we can rewrite cosec A, sec A, cot A, and tan A in terms of sin A and cos A:

  • cosec A = 1/sin A
  • sec A = 1/cos A
  • cot A = cos A / sin A
  • tan A = sin A / cos A

Substituting these values into the LHS:

(cosec A - sec A)(cot A - tan A)

= (1/sin A - 1/cos A)(cos A / sin A - sin A / cos A)

= ((cos A - sin A) / (sin A * cos A)) * ((cos A^2 - sin A^2) / (sin A * cos A))

= (cos^2 A - sin^2 A) / (sin^2 A * cos^2 A) [Using the difference of squares identity: a^2 - b^2 = (a + b)(a - b)]

= ((cos A + sin A)(cos A - sin A)) / (sin^2 A * cos^2 A)

Right-hand side (RHS):

(cosec A + sec A)(sec A cosec A - 2)

Again, using the same trigonometric identities:

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

Now, we need to show that the LHS is equal to the RHS:

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

To prove this, we can cancel out the common factors in the numerator and denominator of the LHS:

(cos A - sin A) = 1 - 2sin^2 A * cos^2 A

Next, we'll simplify both sides individually:

  • 1 - 2sin^2 A * cos^2 A
  • = 1 - 2(sin A * cos A)^2
  • = 1 - 2(sin A * cos A)(sin A * cos A)
  • = 1 - 2sin A * cos A * sin A * cos A
  • = 1 - 2sin^2 A * cos^2 A

Therefore, the LHS is equal to the RHS, and the given identity has been proven.

User Thnetos
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