How do you verify the identity $sin (A + B) = \frac{tan A + tan B}{sec A sec B}$ $sin(A+B)=(tanA+tanB)/(secAsecB)$ ?

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Answer 1 · 2016-12-01T09:04:56

$sin (A + B) = sin A cos B + cos A sin B = \frac{tan A + tan B}{sec A sec B}$ $sin(A+B)=sinAcosB+cosAsinB=(tanA+tanB)/(secAsecB)$

You know that

$sin (A + B) = sin A cos B + cos A sin B$ $sin(A+B)=sinAcosB+cosAsinB$

and, since

$tan α = \frac{sin α}{cos α}$ $tanalpha=sinalpha/cosalpha$

and $sec α = \frac{1}{cos α}$ $secalpha=1/cosalpha$ ,

then

$\frac{tan A + tan B}{sec A sec B} =$ $(tanA+tanB)/(secAsecB)=$

$= \frac{\frac{sin A}{cos A} + \frac{sin B}{cos B}}{\frac{1}{cos A} \cdot \frac{1}{cos B}}$ $=(sinA/cosA+sinB/cosB)/(1/cosA*1/cosB)$

$=((sinAcosB+sinBcosA)/(cancel(cosA)cancelcosB))/(1/(cancelcosAcancelcosB))$

$sinAcosB+sinBcosA$

How do you verify the identity sin(A+B)=(tanA+tanB)/(secAsecB)sin(A+B)=tanA+tanBsecAsecBsin(A+B)=(tanA+tanB)/(secAsecB)?