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.