Question

Please help me to prove this. ​

Answer 1

Answer :

For triangle : A + B + C = π

⇒ A + B = π - C

⇒ cot(A + B) = cot(π - C)

⇒ $\sf\dfrac{cotA\:cotB-1}{cotB+cotA}=-cotA$

⇒ $\bf{cotA\:cotB+cotB\:cotC+cotC\:cotA=1}$

Now 1st part of the given expression!

⇒ $\sf\dfrac{cosA}{sinB\:sinC}$

⇒ $\sf\dfrac{cos(\pi-(B+C))}{sinB\:sinC}$

⇒ $\sf\dfrac{-cosB\:cosC+sinB\:sinC}{sinB\:sinC}$

⇒ 1 - cotB cotC

Similarly 2nd part!

⇒ 1 - cotA cotB

Similarly 3rd part!

⇒ 1 - cotC cotA

LHS :

$\circ\:\sf{3-(cotA\:cotB+cotB\:cotC+cotC\:cotA)}$

$\circ\:\sf{3-1}$

$\circ\:\bf{2}$ = RHS

Hence Proved!!

Answer 2

Answer:  see proof below

Step-by-step explanation:

Given: A + B + C = π                 →  A + B = π - C

                                                 →  B + C = π - A

                                                 →  A + C = π - B

Use the following Double Angle Identity:    sin 2A = 2 sin A · cos A

Use the following Cofunction Identity:    sin A = cos (π/2 - A)

Use the following Sum to Product Identity:

                                                       sin A + sin B = sin [(A + B)/2] · cos [(A - B)/2]

Proof LHS → RHS

$\text{LHS:}\qquad \dfrac{\cos A}{\sin B\cdot \sin C}+\dfrac{\cos B}{\sin C\cdot \sin A}+\dfrac{\cos C}{\sin A\cdot \sin B}\\\\\\.\qquad \quad = \bigg(\dfrac{2\sin A}{2\sin A}\bigg)\dfrac{\cos A}{\sin B\cdot \sin C}+\bigg(\dfrac{2\sin B}{2\sin B}\bigg)\dfrac{\cos B}{\sin C\cdot \sin A}+\bigg(\dfrac{2\sin C}{2\sin C}\bigg)\dfrac{\cos C}{\sin A\cdot \sin B}\\\\\\.\qquad \quad =\dfrac{2\sin A\cdot \cos A+2\sin B\cdot \cos B+2\sin C\cdot \cos C}{2\sin A\cdot \sin B\cdot \sin C}$

$\text{Double Angle:}\qquad \qquad \dfrac{\sin 2A+\sin 2B+\sin 2C}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{(\sin 2A+\sin 2B)+\sin 2C}{2\sin A\cdot \sin B\cdot \sin C}$

$\text{Sum to Product:}\qquad \dfrac{2\sin (A+B)\cdot \cos (A-B)+\sin 2C}{2\sin A\cdot \sin B\cdot \sin C}$

$\text{Given:}\qquad \qquad \qquad \dfrac{2\sin (\pi -C)\cdot \cos (A-B)+\sin 2C}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{2\sin C\cdot \cos (A-B)+\sin 2C}{2\sin A\cdot \sin B\cdot \sin C}$

$\text{Double Angle:}\qquad \qquad \dfrac{2\sin C\cdot \cos (A-B)+2\sin C\cdot \cos C}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{2\sin C(\cos (A-B)+\cos C)}{2\sin A\cdot \sin B\cdot \sin C}$

$\text{Sum to Product:}\qquad \dfrac{2\sin C(2\cos (\frac{A-B+C}{2})\cdot \cos (\frac{A-B-C}{2})}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{4\sin C\cdot \cos (\frac{A+C}{2}-\frac{B}{2})\codt \cos (\frac{A}{2}-\frac{B+C}{2})}{2\sin A\cdot \sin B\cdot \sin C}$

$\text{Given:}\qquad \qquad \dfrac{4\sin C(\frac{\pi-B}{2}-\frac{B}{2})\cdot \cos (\frac{A}{2}-\frac{\pi -A}{2})}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad =\dfrac{4\sin C(\frac{\pi}{2}-B)\cdot \cos (\frac{\pi}{2}-A)}{2\sin A\cdot \sin B\cdot \sin C}$

$\text{Cofunction:}\qquad \qquad \dfrac{4\sin C\cdot \sin B\sin A}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad \qquad =2$

LHS = RHS:  2 = 2  $\checkmark$