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Prove that Sin70∘/Cos20∘ /Cosec20/Sec70 - 2cos20∘Cosec20 = 0

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Question

Prove that `sin 70^@/cos 20^@  + (cosec 20^@)/sec 70^@  -  2 cos 20^@ cosec 20^@ = 0`

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Solution

Sin (70°) = sin (90° - 20°) = cos 20°

Cos 70° = cos (90° - 20°) = sin 20°

`=> cos 20^@/cos 20^@ + sec 70^@/sec 70^@  - 2 sin 20 cosec 20^@`

1 + 1 - 2(1) = 0

∴ LHS = RHS Hence proved

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