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cos13+sin13/cos13-sin13 = tanA find A=?
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two equal signs?
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\[\frac{\cos 13 + \sin 13}{\cos 13 - \sin 13} = \tan A?\]
I think there is one equal sign.. next sign is for finding..
If we are just asked to find A then: \[A = \tan^{-1} (\frac{\cos13 + \sin13}{\cos13 - \sin13})\]
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i think it's using trig identities...
otherwise that would be too easy...just a job for a calculator lol
without the use of calculator
\[\frac{tanA + tanB}{1 - tanA .tanB} = \tan(A + B)\]
\[A = \tan^{-1} (\frac{1 + \tan13}{1 - \tan13})\] \[A = \tan^{-1} (\frac{\tan45 + \tan13}{1 - \tan45.\tan13}) \implies \tan^{-1}(\tan(45 + 13)) \implies A = 58\]
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thankyou
Welcome dear..
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