Question

$\tan(( \frac{\pi}{4} ) + thita) = \frac{ \cos(thita) + \sin(thita) }{ \cos(thita) - \sin(thita) }$
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Answer 1

Answer:

$\huge \star \red{ \underline{question}}$

$\tan( \frac{\pi}{4} + \theta) = \frac{ \cos\theta + \sin\theta }{ \cos\theta - \sin\theta }$

$\huge \green{\underline{ solution}} \\ \\ \\using \: identities \: \implies \\ \red{ \star} \: tan \: \frac{ \pi}{4} = 1 \\ \red{ \star} \: tan \theta = \frac{sin \theta}{cos \theta} \\ \red{ \star} \: tan(x + y) = \frac{tanx + tany}{1 - tanx.tany} \\ \\ step - \: by \: step - \: explanation \\ \\ \star firstly \: taking \: left \: hand \: side \\ \implies \: tan( \frac{ \pi}{4} + \theta) \\ \\ using \: given \: identity \\ \implies \: \frac{tan \frac{ \pi}{4} + tan \theta}{1 - tan \frac{ \pi}{4}.tan \theta } \\ \\ \implies \: \frac{1 + \frac{sin \theta}{cos \theta} }{1 - \frac{sin \theta}{cos \theta} } \\ \\ \implies \red{taking \: LCM }\\ \\ \implies \frac{ \frac{cos \theta + sin \theta}{cos \theta} }{ \frac{cos \theta - sin \theta}{cos \theta} } \\ \\ \implies \: \frac{cos \theta + sin \theta}{cos \theta - sin \theta} \\ left \: hand \: side \: = right \: hand \: side \: \\ \\ hence\: proved$