Question

Prove this equation ​

Answer 1

sin^2 A-sin^ B. + cos^2A-cos^B

sin^2 A-sin^ B. + cos^2A-cos^B------------------------------------------------------

sin^2 A-sin^ B. + cos^2A-cos^B------------------------------------------------------ (cos A+cos B) (sin A+sin B)

sin^2 A-sin^ B. + cos^2A-cos^B------------------------------------------------------ (cos A+cos B) (sin A+sin B) sin^2 A +cos^A -(sin^B +cos^B)

sin^2 A-sin^ B. + cos^2A-cos^B------------------------------------------------------ (cos A+cos B) (sin A+sin B) sin^2 A +cos^A -(sin^B +cos^B)= -----------------------------------------------------

sin^2 A-sin^ B. + cos^2A-cos^B------------------------------------------------------ (cos A+cos B) (sin A+sin B) sin^2 A +cos^A -(sin^B +cos^B)= ----------------------------------------------------- (cos A+cos B) (sin A+sin B)

sin^2 A-sin^ B. + cos^2A-cos^B------------------------------------------------------ (cos A+cos B) (sin A+sin B) sin^2 A +cos^A -(sin^B +cos^B)= ----------------------------------------------------- (cos A+cos B) (sin A+sin B) 1-1

sin^2 A-sin^ B. + cos^2A-cos^B------------------------------------------------------ (cos A+cos B) (sin A+sin B) sin^2 A +cos^A -(sin^B +cos^B)= ----------------------------------------------------- (cos A+cos B) (sin A+sin B) 1-1=-------------------------------------------------

sin^2 A-sin^ B. + cos^2A-cos^B------------------------------------------------------ (cos A+cos B) (sin A+sin B) sin^2 A +cos^A -(sin^B +cos^B)= ----------------------------------------------------- (cos A+cos B) (sin A+sin B) 1-1=------------------------------------------------- (cos A+cos B) (sin A+sin B)

sin^2 A-sin^ B. + cos^2A-cos^B------------------------------------------------------ (cos A+cos B) (sin A+sin B) sin^2 A +cos^A -(sin^B +cos^B)= ----------------------------------------------------- (cos A+cos B) (sin A+sin B) 1-1=------------------------------------------------- (cos A+cos B) (sin A+sin B) =0 (proved)

Answer 2

To prove:

$\dfrac{sin A - sin B}{cosA + cosB} +\dfrac {cosA - cos B}{sinA + sin B} =0$

Proof:

By taking LCM

$LHS= \dfrac{(sinA - sin B)(sin A + Sin B) + (cos A - cos B)(cos A + cos B)}{(cos A + cos B)(sin A + sin B)}$

$=\dfrac {({sin}^{2} A - {sin}^{2} B) + ({cos}^{2}A - {cos}^{2} B)}{(cos A + cosB)(sin A + sin B)}$

$=\dfrac{{sin}^{2}A - {sin}^{2}B + {cos}^{2}A - {cos}^{2} B}{(cos A + cos B)(sin A + sin B)}$

$=\dfrac {{sin}^{2} A +{cos}^{2}A - ({sin}^{2} B +{cos}^{2}B)}{(cos A + cos B)(sin A + sin B)}$

$=\dfrac{ 1 - 1}{(cos A + cos B)(sin A + sin B)}$

$= \dfrac{0}{(cos A + cos B)(sin A + sin B)}$

= 0 = RHS

Identity used:

${sin}^{2}\theta + {cos}^{2}\theta = 1$

Hence, proved.