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(sin70/cos20)+(cos36/sec54)-(2cos43cos47/tan10tan40tan80tan10)

Asked by sanskriti 19th September 2013, 12:27 PM
Answered by Expert
Answer:

Pls cross check your question, it should be as written below.

(sin70/cos20)+(cosec36/sec54)-(2cos43cosec47/tan10tan40tan50tan10)

= sin[90-20] / [cos20] + [cosec36] / sec[90-36] - {2cos43 cosec [90-43] / tan10. tan40. tan[90-40]. tan[90-10]}

=[cos20] / [cos20] + [cosec36 / cosec36] - [2cos43sec43 / tan10. tan40. cot40. cot10]

= 1 + 1 - 2

= 0

Answered by Expert 19th September 2013, 12:36 PM
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