How do you verify the identity $sec (A - B) = \frac{sec A sec B}{1 + tan A tan B}$ $sec(A-B)=(secAsecB)/(1+tanAtanB)$ ?

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Answer 1 · 2017-06-21T18:12:42

For Proof, refer to the Explanation.

$sec (A - B) = \frac{1}{cos (A - B)} = \frac{1}{cos A cos B + sin A sin B}$ $sec(A-B)=1/cos(A-B)=1/(cosAcosB+sinAsinB)$

$= \frac{1}{cos A cos B (1 + \frac{sin A sin B}{cos A cos B})},$ $=1/{cosAcosB(1+(sinAsinB)/(cosAcosB))},$

$= \frac{1}{cos A cos B (1 + tan A tan B)},$ $=1/{cosAcosB(1+tanAtanB)},$

$= \frac{\frac{1}{cos A cos B}}{1 + tan A tan B},$ $={1/(cosAcosB)}/(1+tanAtanB),$

$\Rightarrow sec (A - B) = (sec A sec B) (1 + tan A tan B) .$ $rArr sec(A-B)=(secAsecB)(1+tanAtanB).$

How do you verify the identity sec(A−B)=secAsecB1+tanAtanBsec(A-B)=(secAsecB)/(1+tanAtanB)?