How to determine of tangent (A – B) and tangent (A + B)

We can use the identity \tan \theta =\frac{\sin \theta}{\cos \theta} and the identities for sin (A + B) and cos (A + B) to write identities for tan (A + B) and tan (A B).

Tangent of (A + B)

\tan (A+B)=\frac{\sin (A+B)}{\cos (A+B)}=\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}

Also Read:

We would like to write this identity for tan (A + B) in terms of tan A and tan B. We can do this by dividing each term of the numerator and each term of the denominator by cos A cos B. When we do this we are dividing by a fraction equal to 1 and therefore leaving the value of the expression unchanged.

Tangent of (A B)

The identity for tan (A B) can be derived in a similar manner.

These identities are true for all replacements of A and B for which \cos A\neq 0 and \cos B\neq 0, and for which tan (A + B) or tan (A B) are defined.

Also Read : Determine sine ( A – B) and sine ( A + B) of trigonometric

Example 1:

Use \tan 2\pi =0 and \tan \frac{\pi}{4}=1 to show that \tan \frac{7\pi}{4}=-1.


\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}

\tan (2\pi-\frac{\pi}{4})=\frac{\tan 2\pi-\tan \frac{\pi}{4}}{1+\tan 2\pi \tan \frac{\pi}{4}}



Example 2:

Use (45° + 120°) = 165° to find the exact value of tan 165°.


\sin 45^0=\frac{\sqrt{2}}{2} and \cos 45^0=\frac{\sqrt{2}}{2}, so \tan 45^0=1.

\sin 120^0=\frac{\sqrt{3}}{2} and \cos 120^0=-\frac{1}{2}, so \tan 120^0=-\sqrt{3}.

\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}

\tan (45+120)=\frac{\tan 45+\tan 120}{1-\tan 45\tan 120}


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