Math, asked by ashats90499, 1 year ago

Tan(π÷4 + A÷2) = 1÷secA- tanA

Answers

Answered by rohitkumargupta
4

HELLO DEAR,



TO PROVE :-


Tan(π/4 + A/2) = 1 /( secA - tanA)



WE KNOW:-


sin²A + cos²A = 1,


cos²A - sin²A = cos2A


2sinAcosA = sin2A



\bold{tan(A + B) = \frac{tanA + tanB}{1 - tanAtanB}}



so, \bold{tan(\Pi/4 + A/2) = \frac{tan\Pi/4 + tanA/2}{1 - tan\Pi/4*tanA/2}}



[tanπ/4 = 1]



\bold{\Rightarrow tan(\Pi/4 + A/2) = \frac{1 + tanA/2}{1 - tanA/2}}



\bold{\Rightarrow \frac{1 + \frac{sinA/2}{cosA/2}}{1 - \frac{sinA/2}{cosA/2}}}



\bold{\Rightarrow \frac{sinA/2 + cosA/2}{cosA/2 - sinA/2}}



\bold{\Rightarrow \frac{sinA/2 + cosA/2}{sinA/2 - cosA/2} * \frac{cosA/2 + sinA/2}{cosA/2 + sinA/2}}



\bold{\Rightarrow \frac{(sinA/2 + cosA/2)^2}{(cos^2A/2 - sin^2A/2)}}



\bold{\Rightarrow \frac{(sin^2A/2 + cos^2A/2) + 2sinA/2*cosA/2}{cos2*(\frac{A}{2})}}



\bold{\Rightarrow \frac{1 + sinA}{cosA}}



\bold{\Rightarrow} secA + tanA



\bold{\Rightarrow \frac{secA + tanA}{sec^2A - tan^2A}}


\bold{\to\to\to\to\to\to\to\to\to\to\to\boxed{sin^2\Theta - tan^2\Theta = 1}}



\bold{\Rightarrow \frac{1}{secA - tanA} * \frac{secA + tanA}{secA + tanA}}



\bold{\therefore, \Rightarrow \frac{1}{secA - tanA}}



\mathbb{\ulcorner \star HENCE,\, L.H.S = R.H.S \star \urcorner}



I HOPE ITS YOU DEAR,


THANKS

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