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Question
Prove the following identities:
`(cot A - cos A)/(cot A + cos A) = ("cosec" A - 1)/("cosec" A + 1)`
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Solution
Given: cot A – cos A ---------- = ? cot A + cos A
To Prove: `(cot A - cos A)/(cot A + cos A) = ("cosec" A - 1)/("cosec" A + 1)`.
Proof [Step-wise]:
1. Write cot and cosec in terms of sin and cos:
`cot A = (cos A)/(sin A)`
`"cosec" A = 1/(sin A)`
2. Start with LHS and substitute:
LHS = `(cot A - cos A) / (cot A + cos A)`
= `((cos A/sin A) - cos A) /((cos A/sin A) + cos A)`
3. Factor cos A from numerator and denominator:
LHS = `(cos A (1/sin A - 1))/(cos A (1/sin A + 1))`
4. Cancel the common factor cos A (allowed where cos A ≠ 0):
LHS = `(1/sin A - 1)/(1/sin A + 1)`
5. Replace `1/(sin A)` by cosec A:
LHS = `("cosec" A - 1)/("cosec" A + 1)` = RHS
Hence LHS = RHS.
Therefore `(cot A - cos A)/(cot A + cos A) = ("cosec" A - 1)/("cosec" A + 1)`, for all A where the expressions are defined.
Notes
All steps used division by sin A and cancellation of cos A, so the identity holds for angles A with sin A ≠ 0 and cos A ≠ 0 (equivalently A ≠ nπ and `A ≠ π/2 + nπ`, n integer), i.e. where both sides of the original equality are defined.
