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

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

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Answer 1 · 2015-05-08T14:03:50

Right term, numerator:

$R N = \frac{sin A}{cos A} + \frac{sin B}{cos B} = \frac{sin A . cos B + sin B . cos A}{cos A . cos B}$ $RN = sin A/cos A + sin B/cos B = (sin A.cos B + sin B.cos A)/(cos A.cos B)$

Right term denominator:

$R D = \frac{1}{cos A} (\frac{1}{cos B}) = \frac{1}{cos A . cos B}$ $RD = 1/cos A(1/cos B) = 1/(cos A.cos B)$

$\frac{R N}{R D} = (sin A . cos B + sin B . cos A) = sin (A + B)$ $(RN)/(RD) = (sin A.cos B + sin B.cos A) = sin (A + B)$ Correct.

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

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

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