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

Question

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

1 Answer

Impact of this question

5485 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-16T03:02:06

$cos A cos B - sin A sin B = cos A cos B - sin A sin B$ $cosAcosB-sinAsinB=cosAcosB-sinAsinB$

Explanation:

$cos (A + B) = cos A cos B - sin A sin B$ $cos(A+B)=cosAcosB-sinAsinB$

$\frac{1 - tan A tan B}{sec A sec B} = (\frac{1}{sec A}) (\frac{1}{sec B}) - \frac{tan A tan B}{sec A sec B}$ $(1-tanAtanB)/(secAsecB)=(1/secA)(1/secB)-(tanAtanB)/(secAsecB)$

$(\frac{1}{sec A}) (\frac{1}{sec B}) = cos A cos B$ $(1/secA)(1/secB)=cosAcosB$

$(\frac{tan A tan B}{sec A sec B}) = \frac{(\frac{sin A}{cos A}) (\frac{sin B}{cos B})}{(\frac{1}{cos A}) (\frac{1}{cos B})} = sin A sin B$ $((tanAtanB)/(secAsecB))=((sinA/cosA)(sinB/cosB))/((1/cosA)(1/cosB))=sinAsinB$

$cos A cos B - sin A sin B = cos A cos B - sin A sin B$ $cosAcosB-sinAsinB=cosAcosB-sinAsinB$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you verify the identity cos(A+B)=(1-tanAtanB)/(secAsecB)cos(A+B)=1−tanAtanBsecAsecBcos(A+B)=(1-tanAtanB)/(secAsecB)?

1 Answer

Explanation:

Related questions

Impact of this question

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