Question

In triangle ABC Prove that,
$\sf cos(A+B)+sinC=sin(A+B)-cosC$​

Answer 1

Question : -

In ∆ABC, Prove that;

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

ANSWER

Given : -

ABC is a ∆

Required to prove : -

• cos (A+B) + sin C = sin (A+B) - cos C

Proof : -

Given that;

∆ABC is a ∆ where angles are A,B,C

so,

A + B + C = 180° (sum angle property)

Now,

A + B = 180° - C

By taking sin function on both sides

sin (A + B) = sin (180° - C)

sin (A + B) = sin C »»»(1)

This is because in the 2nd quadrant sin function is positive .

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Now, instead of sin let's take cos

cos (A + B) = cos (180° - C)

cos (A + B) = - cos C »»»(2)

This is because cos function is negative in the π quadrant .

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To prove !

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

consider LHS part !

Consider the LHS part !

cos (A+B) + sin C

From (2)

- cos C + sin C

=> sin C - cos C ..(3)

Consider the RHS part !

sin (A+B) - cos C

From (1)

sin C - cos C ..(4)

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From (3)&(4) we have;

LHS = RHS = sin C - cos C

Hence Proved !

Answer 2

QuestioN :-

In a triangle ABC , Prove that :-

• $\sf \: cos(a + b) + sin \: c = sin \: (a + b) - cos \: c$

SolutioN :-

Given that ABC is a triangle and we know that The sum of all three angle in a triangle = 180

So,

$\star \sf \: A + B + C = 180 \\ \\ \star \sf \: A + B = 180 - C - - - - (i) \\ \\ </u><u>$

Now,

$\sf \: cos(a + b) + sin \: c = sin(a + b) - cos \: c$

Now substituting the value of (A + B) from equation (i) :-

$\implies \sf \: cos(180 - c) + sin \: c = sin(180 - c) - cos \: c$

We know that :-

• $\sf \: cos(180 + c) = - cos \: c$
• $\sf \: sin(180 - c) = sin \: c$

So,

$\implies \sf \: - cos \: c + sin \: c = sin \: c - cos \: c \\ \\ \implies \sf \: \: \sf \: - cos c + cos c = sin c - sin c \\ \\ \implies \sf \: \: 0 = 0$

LHS = RHS

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Hope it will help you :)