Conditional Trigonometric Identities - Practice Questions & MCQ

Conditional Identities

Till now we have come across many trigonometric identities, such as $\sin ^2 \theta+\cos ^2 \theta=1, \sec ^2 \theta$ $\tan ^2 \theta=1$ etc. Such identities are true for all the angles which are in the domain. In this section, we are going to learn some conditional identities.

Here the condition is that $A, B$, and $C$ are the angles of triangle $A B C$, and $A+B+C=\pi$.
As, $\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi$ then, $\mathrm{A}+\mathrm{B}=\pi-\mathrm{C}, \mathrm{A}+\mathrm{C}=\pi-\mathrm{B}$ and $\mathrm{B}+\mathrm{C}=\pi-\mathrm{A}$
Using the above conditions, we can get some important identities.
1. $\sin (A+B)=\sin (\pi-C)=\sin C$

Similarly, $\sin (A+C)=\sin B$ and $\sin (B+C)=\sin A$
2. $\cos (A+B)=\cos (\pi-C)=-\cos C$

Similarly, $\cos (A+C)=-\cos B$ and $\cos (B+C)=-\cos A$
3. $\tan (A+B)=\tan (\pi-C)=-\tan C$

Similarly, $\tan (A+C)=-\tan B$ and $\tan (B+C)=-\tan A$

Example 1

$
\tan A+\tan B+\tan C=\tan A \cdot \tan B \cdot \tan C, \text { where } \mathrm{A}+\mathrm{B}+\mathrm{c}=\pi
$
Proof:

$
\begin{aligned}
& \Rightarrow A+B=\pi-C \\
\Rightarrow & \tan (A+B)=\tan (\pi-C) \\
\Rightarrow & \frac{\tan A+\tan B}{1-\tan A \tan B}=-\tan C \\
\Rightarrow & \tan A+\tan B=-\tan C+\tan A \tan B \tan C \\
\Rightarrow & \tan A+\tan B+\tan C=\tan A \tan B \tan C
\end{aligned}
$
Example 2

$
\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{C}{2} \tan \frac{B}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1
$
Proof:
since $A+B+C=\pi$, we have $\frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}$

$
\begin{aligned}
& \Rightarrow \tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right)=\cot \frac{C}{2} \\
& \Rightarrow \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}} \\
& \Rightarrow \tan \frac{A}{2} \tan \frac{C}{2}+\tan \frac{B}{2} \tan \frac{C}{2}=1-\tan \frac{A}{2} \tan \frac{B}{2} \\
& \Rightarrow \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1
\end{aligned}
$