Question

1+cos thita÷1-cos thita - 1-cos thita÷1+cos thita

Answer 1

Answer:

4cotAcosecA

Explanation:

Here I am using A instead of theta.

(1+cosA)/(1-cosA)-(1-cosA)/1+cosA)

= [(1+cosA)²-(1-cosA)²]/[(1+cosA)(1-cosA)]

= ( 4cosA)/[1²-cos²A]

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Since ,

i )(a+b)²-(a-b)² = 4ab

ii ) (a+b)(a-b) = a²-b²

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= (4cosA)/sin²A

= 4(cosA/sinA)×(1/sinA)

= 4cotAcosecA.

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