Question

a b c are the angle of a triangle then prov that. sin(A+B)-sinC=0
​

Answer 1

Answer:

Proved below.

Step-by-step explanation:

We know that

The sum of all angles in triangle = 180°

⇒ A + B + C = 180°

⇒ A + B = 180 - C

Take 'sin' on both sides

⇒ sin( A+B ) = sin ( 180 - C )

⇒ sin( A+B ) = sin C                 [∵sin (180-θ) = sin θ]

∴ sin( A+B ) - sin C = 0

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• If A, B, C are the angles of triangle ABC, prove that sin(A + B) - sinC = 0

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\purple{\bold{Solution :- }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

☆ We know,

☆ In triangle ABC,

$\sf \:  ⟼∠A + ∠B + ∠C = 180°$

$\sf \:  ⟼∠A + ∠B = 180° - ∠C$

$\sf \:  ⟼Now, sin(A + B) = sin(180° - C) = sinC$

$\sf \:  (Because, \: In \: second \: quadrant, \: sin \: is \: positive.)$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\bf \:  ⟼Consider, \: LHS$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\bf \:sin(A + B) - sinC$

$\bf \: = sinC - sinC$

$\bf \: = \cancel{sinC} - \cancel{sinC}$

$\bf \: = 0$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large \red{\bf \: ⟼ Explore \: more } ✍$

─━─━─━─━─━─━─━─━─━─━─━─━─

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

─━─━─━─━─━─━─━─━─━─━─━─━─

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

─━─━─━─━─━─━─━─━─━─━─━─━─

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

─━─━─━─━─━─━─━─━─━─━─━─━─

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

─━─━─━─━─━─━─━─━─━─━─━─━─

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)

─━─━─━─━─━─━─━─━─━─━─━─━─

Sign of Trigonometric ratios in Quadrants

sin (90°-θ)  =  cos θ

cos (90°-θ)  =  sin θ

tan (90°-θ)  =  cot θ

csc (90°-θ)  =  sec θ

sec (90°-θ)  =  csc θ

cot (90°-θ)  =  tan θ

sin (90°+θ)  =  cos θ

cos (90°+θ)  =  -sin θ

tan (90°+θ)  =  -cot θ

csc (90°+θ)  =  sec θ

sec (90°+θ)  =  -csc θ

cot (90°+θ)  =  -tan θ

sin (180°-θ)  =  sin θ

cos (180°-θ)  =  -cos θ

tan (180°-θ)  =  -tan θ

csc (180°-θ)  =  csc θ

sec (180°-θ)  =  -sec θ

cot (180°-θ)  =  -cot θ

sin (180°+θ)  =  -sin θ

cos (180°+θ)  =  -cos θ

tan (180°+θ)  =  tan θ

csc (180°+θ)  =  -csc θ

sec (180°+θ)  =  -sec θ

cot (180°+θ)  =  cot θ

sin (270°-θ)  =  -cos θ

cos (270°-θ)  =  -sin θ

tan (270°-θ)  =  cot θ

csc (270°-θ)  =  -sec θ

sec (270°-θ)  =  -csc θ

cot (270°-θ)  =  tan θ

sin (270°+θ)  =  -cos θ

cos (270°+θ)  =  sin θ

tan (270°+θ)  =  -cot θ

csc (270°+θ)  =  -sec θ

sec (270°+θ)  =  cos θ

cot (270°+θ)  =  -tan θ