Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
1 answer:
Answer:
The answer is below
Step-by-step explanation:
We need to prove that:
(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.
Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.
Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

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I am sorry but please give detailed question