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If A,B, and C is the angles of triangles then Simplify the function \sin(A+B)\cos(A+B)+ \sin(B+C)\cos(B+C) + \sin(C+A)\cos(C+A)

Option: 1

2 \sin A \cdot \sin B\cdot \sin C


Option: 2

\sin A \cdot \sin B\cdot \sin C


Option: 3

4 \sin A \cdot \sin B\cdot \sin C


Option: 4

2 \sin A + 2\sin B +2 \sin C


Answers (1)

best_answer

Conditional Identities -

Conditional Identities

 

Till now we have come across many trigonometric identities, such as sin2? + cos2? = 1, sec2? - tan2? = 1 etc. Such identities are true for all the angles which satisfy the given conditions. 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 ABC, and A + B + C = π.

As, A + B + C = π then,  A + B = π - C, A + C = π - B and  B + C = π - A

   Using the above conditions, we can get some important identities.

  1. sin (A + B) = sin (π - C) = sin C

Similarly, sin (A + C) = sin B and sin (B + C) = sin A

  1. cos (A + B) = cos (π - C) = -cos C

Similarly, cos (A + C) = -cos B and cos (B + C) = -cos A

  1. tan (A + B) = tan (π - C) = -tan C

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

Example

\tan A+\tan B+\tan C=\tan A \cdot\tan B \cdot\tan C, Where A + B + C = π

Where A + B + C = π

Proof:

{\Rightarrow A+B=\pi-C} \\ {\Rightarrow \tan (A+B)=\tan (\pi-C)} \\ {\Rightarrow \quad \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}

Example

\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{aligned} & \text { since } A+B+C=\pi, \text { we have } \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2} \\ \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}
 

 

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\text{Given A, B, C is the angles of triangles so A+B+C}=\pi \\ \sin(A+B)\cos(A+B)+ \sin(B+C)\cos(B+C) + \sin(C+A)\cos(C+A)=-\sin(C)\cos(C)-\sin(A)\cos(A)-\sin(B)\cos(B)\\ \ \ \ \{ \text{if A+B+C} =\pi \ \ \ then\\sin(A+B)=\sin(C),and\ \cos(A+B)=-\cos(C) \} \\ \sin(A+B)\cos(A+B)+ \sin(B+C)\cos(B+C) + \sin(C+A)\cos(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

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rishi.raj

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