`tan(pi/4 - theta) = (1 - tan(theta))/(1 + tan(theta))` Prove the identity.

$\displaystyle{\tan{{\left(\frac{\pi}{{4}}-\theta\right)}}}=\frac{{{1}-{\tan{{\left(\theta\right)}}}}}{{{1}+{\tan{{\left(\theta\right)}}}}}$

we will use the following formula to prove the identity,

$\displaystyle{\tan{{\left({A}-{B}\right)}}}=\frac{{{\tan{{A}}}-{\tan{{B}}}}}{{{1}+{\tan{{A}}}{\tan{{B}}}}}$

LHS= $\displaystyle{\tan{{\left(\frac{\pi}{{4}}-\theta\right)}}}$

$\displaystyle=\frac{{{\tan{{\left(\frac{\pi}{{4}}\right)}}}-{\tan{{\left(\theta\right)}}}}}{{{1}+{\tan{{\left(\frac{\pi}{{4}}\right)}}}{\tan{{\left(\theta\right)}}}}}$

plug in the value of tan(pi/4)=1,

$\displaystyle=\frac{{{1}-{\tan{{\left(\theta\right)}}}}}{{{1}+{\tan{{\left(\theta\right)}}}}}$

LHS=RHS, Hence proved.

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Monday, December 30, 2013

$\displaystyle{\tan{{\left(\frac{\pi}{{4}}-\theta\right)}}}=\frac{{{1}-{\tan{{\left(\theta\right)}}}}}{{{1}+{\tan{{\left(\theta\right)}}}}}$ Prove the identity.

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