Conditional Trigonometric Identities

Conditional trigonometric identities come into play whenever trigonometric functions appear in an expression or equation. They hold true for every possible value of the variables involved on both sides of the equation. Geometrically, these identities relate to specific trigonometric functions like sine, cosine, and tangent, which involve one or more angles.

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Conditional Identities

Trigonometric identities are equations involving trigonometric functions that remain true for all values of the variables in the equation. These identities involve various relationships between the sides and angles of a triangle, particularly in the context of right-angle triangles. They rely on the fundamental trigonometric ratios: sine, cosine, tangent, cosecant, secant, and cotangent. These ratios are defined based on the sides of a right triangle — the adjacent side, opposite side, and hypotenuse.

Fundamental trigonometric identities are derived directly from these ratios, forming the basis for solving trigonometric problems across mathematics and science.

Till now we have come across many trigonometric identities, such as ${\sin }^2\theta +{\cos }^2\mathrm{\Theta }=1,{\sec }^2\mathrm{\Theta }-{\tan }^2\theta =1$ 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 $\mathrm{A},\mathrm{B}$, and C are the angles of triangle ABC , 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

$\tan A+\tan B+\tan C=\tan A\cdot \tan B\cdot \tan C$, where $\mathrm{A}+\mathrm{B}+$ $C=\pi$

Proof:

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

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:

$\begin{array}{rl}& \text{since}A+B+C=\pi \text{, we have}\frac{A}{2}+\frac{B}{2}=\frac{\pi }{2}-\frac{C}{2}\\ \Rightarrow & \tan (\frac{A}{2}+\frac{B}{2})=\tan (\frac{\pi }{2}-\frac{C}{2})=\cot \frac{C}{2}\\ \Rightarrow & \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}\tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}}\\ \Rightarrow & \tan \frac{A}{2}\tan \frac{C}{2}+\tan \frac{B}{2}\tan \frac{C}{2}=1-\tan \frac{A}{2}\tan \frac{B}{2}\\ \Rightarrow & \tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=1\end{array}$

Summary

Conditional identities are crucial tools that help us understand and use trigonometry effectively. They simplify complex calculations, making it easier to apply trigonometry in mathematics and science. These identities play a key role in advancing our knowledge and capabilities in these fields.

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Solved Examples Based on Conditional Identities

Example 1: If $A,B$, and $C$ is the angles of triangles then Simplity the function

$\sin (A+B)\cos (A+B)+\sin (B+C)\cos (B+C)+\sin (C+A)$

1) $2\sin A\cdot \sin B\cdot \sin C$
2) $\sin A\cdot \sin B\cdot \sin C$
3) $4\sin A\cdot \sin B\cdot \sin C$
4) $2\sin A+2\sin B+2\sin C$

Solution

Given $\mathrm{A},\mathrm{B},\mathrm{C}$ is the angles of triangles so $\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi$

$\begin{array}{rl}& \sin (A+B)\cos (A+B)+\sin (B+C)\cos (B+C)+\sin (C+A\\ & -\sin (C)\cos (C)-\sin (A)\cos (A)-\sin (B)\cos (B)\\ & \{\text{if}\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \text{then}\\ & \sin (A+B)=\sin (C)\text{, and}\cos (A+B)=-\cos (C)\}\\ & \sin (A+B)\cos (A+B)+\sin (B+C)\cos (B+C)+\sin (C+A\\ & \frac{1}{2}\{2\sin (C)\cos (C)+2\sin (A)\cos (A)+2\sin (B)\cos (B)\}\\ & =\frac{1}{2}\{\sin 2C+\sin 2A+\sin 2B\}\\ & =\frac{1}{2}\{4\sin A\sin B\sin C\}\{\text{by above concept}\}\\ & =2\sin A\sin B\sin C\end{array}$
Hence, the answer is option 4.

Example 2: If $\mathrm{A},\mathrm{B}$, and C is the angles of a triangle, and $\tan A+\tan B+\tan C=3\sqrt{3}$, then which of the following is true?
1) Three different triplets of angles are possible
2) $ABC$ is a right triangle
3) $ABC$ can be an equilateral triangle but not necessarily
4) $ABC$ is an equilateral triangle

Solution
Given $\tan A+\tan B+\tan C=3\sqrt{3}$
we know

$\begin{array}{rl}& \tan A+\tan B+\tan C=\tan A\tan B\tan C\{\text{for triangle}\mathrm{ABC}\}\\ & A\cdot M\cdot \ge G\cdot M\\ & \frac{\tan A+\tan B+\tan C}{3}\ge (\tan A\tan B\tan C{)}^{\frac{1}{3}}\\ & \sqrt{3}\ge (3\sqrt{3}{)}^{\frac{1}{3}}\\ & A\cdot M.=G\cdot M.\text{only possible when}\tan A=\tan B=\tan C\text{or}\mathrm{A}=\mathrm{B}=\mathrm{C}\\ & A+B+C=\pi \\ & \text{so}A=B=C=\frac{\pi }{3}\end{array}$

Hence, the answer is option 4.

Example 3: Number of solutions of $\tan x+\tan 3x+\tan 5x=\tan x\tan 3x\tan 5x$ for $x\in [0,\frac{\pi }{2}]$ is?
1) 3
2) 4
3) 5
4) 6

Solution

Conditional Identities -
Here the condition is that $A,B$, and $C$ are the angles of triangle $ABC$, and $A+B+C=\pi$.
As, $A+B+C=\pi$ then, $A+B=\pi -C,A+C=\pi -B$ and $B+C=\pi -A$
Using the above conditions, we can get some important identities.

$\text{1.}\sin (A+B)=\sin (\pi -C)=\sin C$
Similarly, $\sin (A+C)=\sin B$ and $\sin (B+C)=\sin A$

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

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

$\begin{array}{rl}& \tan x+\tan 3x+\tan 5x=\tan x\tan 3x\tan 5x\text{is possible when}\mathrm{x}+3\mathrm{x}+5\mathrm{x}=n\pi \\ & x=\frac{n\pi }{9}\\ & x=\{\frac{\pi }{9},\frac{2\pi }{9},\frac{\pi }{3},\frac{4\pi }{9}\}\\ & \text{Number of solution is}4\end{array}$
Example 4: If $\sin 2\mathrm{A}+\sin 2\mathrm{B}+\sin 2\mathrm{C}=\alpha \sin \mathrm{A}\sin \mathrm{B}\sin \mathrm{C}$ then the value of $\alpha$ is?
(Given $\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi$ )
1) 2
2) 4
3) 6
4) 1

Solution
We have to find the value of $\alpha$ such that $\sin 2A+\sin 2B+\sin 2C=\alpha \sin A\sin B\sin C$

$\begin{array}{rl}(\sin 2A+\sin 2B)+\sin 2C& =2\sin (A+B)\cos (A-B)+\sin 2C\\ & =2\sin (\pi -C)\cos (A-B)+\sin 2C\\ & =2\sin C\cos (A-B)+2\sin C\cos C\\ & =2\sin C[\cos (A-B)+\cos C]\\ & =2\sin C[\cos (A-B)+\cos C]\\ & =2\sin C[\cos (A-B)-\cos (A+B)]\\ & =2\sin C\times 2\sin A\sin B=4\sin A\sin B\sin C\end{array}$

Therefore, the value of $\alpha$ is 4.

Example 5: If ${\sin }^{-1}a+{\sin }^{-1}b+{\sin }^{-1}c=\pi$, then the value of $a\sqrt{(1-a^2)}+b\sqrt{(1-b^2)}+c\sqrt{(1-c^2)}$ will be
1) $2abc$
2) $abc$
3) $\frac{1}{2}abc$
4) $\frac{1}{3}abc$

Solution
Double Angle Formula -

$\begin{array}{rl}\sin 2\alpha & =2\sin \alpha \cos \alpha \\ \cos 2\alpha & ={\cos }^2\alpha -{\sin }^2\alpha \\ & =2{\cos }^2\alpha -1\\ & =1-2{\sin }^2\alpha \\ \tan 2\alpha & =\frac{2\tan \alpha }{1-{\tan }^2\alpha }\end{array}$

- wherein

These are formulae for double angles
$\begin{array}{rl}& \text{Let}{\sin }^{-1}a=A\\ & {\sin }^{-1}b=B\\ & {\sin }^{-1}c=C\\ & \therefore \sin A=a,\sin B=b,\sin C=c\end{array}$

and $A+B+C=\pi$, then

$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$
Now $a\sqrt{(1-a^2)}+b\sqrt{(1-c^2)}+c\sqrt{(1-c^2)}$

$\begin{array}{rl}& =\sin A\cos A+\sin B\cos B+\sin C\cos C\\ & =\frac{1}{2}[\sin 2A+\sin 2B+\sin 2C]=2\sin A\sin B\sin C=2abc\end{array}$
Maths No Difficulty Level $6\underset{―}{170850}$
In $\mathrm{△}ABC$ tan $A+\tan B+\tan C$ is equal to.