Math, asked by barbiesophia123, 1 year ago

plz...answer this question

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Answered by shanujindal48p68s3s

$\frac{ \tan(50) + \sec(50) }{ \cot(40) + \csc(40) } + \cos(40) \csc(50) \\ = \frac{ \cot(90 - 50) + \ \csc (90 - 50) }{ \cot(40) + \csc(40) } + \cos(40) \sec(90 - 50) \\ = 1 + \frac{ \cos(40) }{ \cos(40) } \\ = 1 + 1 = 2$
Brainliest please!!

littledora: plz...can u explain me this answer I am not understanding

shanujindal48p68s3s: We know that sin x= cos (90-x) and tan x = cot (90-x) and sec x = cosec(90-x) and vice versa

littledora: ok

shanujindal48p68s3s: Brainliest?

shanujindal48p68s3s: Thanks!!

littledora: welcome

Answered by littledora

this is helpful for u

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shanujindal48p68s3s: Dude you need to work on your division, in the last step the first term is cancelled wrong

littledora: ok

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