Question

Prove that :
sin(A+B) =sin A.cos B + cos A.sin B

ICSE class 10th​

Answer 1

To Prove :-

sin ( A + B ) = sinA. cosB + cosA. sinB

Answer :-

In the given figure ,

∠TPR + ∠PRT = 90°

and , ∠ORT + ∠PRT = 90°

•°• ∠TPR = ∠ORT = ∠ROX = A

Now , In right triangle POQ ,

sin ( A + B )

= PQ / OP

= ( PT + TQ ) / OP

= PT / OP + TQ / OP

= PT / OP + RS / OP [ °•° TQ = RS ]

= PT / OP × PR / OP + RS / OR . OR / OP

= cosA. sinB + sinA. cosB [ °•° ∠TPR = A ]

= sinA cosB + cosA sinB. [ Hence Proved ]

Answer 2

ANSWER

This question can be proved in many different ways.

I will do this proof this geometrically.

Now, for diagram plz refer to the attachment given.

We know=>

-Angle B +(A-B) =B+A-B =A

So, Angle PQR = A

-Angle OPS =90°

-length of OS = 1(unit length)

To prove=

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

Proof=>

Sin(A-B) = $\dfrac{O}{H}$

=>Sin(A-B) = $\dfrac{ST}{1}$

=>Sin(A-B) =ST. - - - - - - - - (i)

Now we have =>

ST = PR - PQ

(Since QR = ST)

Now,

-Sin(B) =$\dfrac{O}{H}$

=>Sin(B) =$\dfrac{PS}{1}$

=>Sin(B) = PS

-Cos(B) = $\dfrac{A}{H}$

=>Cos(B) = $\dfrac{OP}{1}$

=>Cos(B) =OP

-Sin(A) = $\dfrac{O}{H}$

=>Sin(A) = $\dfrac{PR}{CosB}$

=>SinACosB = PR - - - - - - (ii)

AND CosASinB = PQ. - - - - - - - (iii)

So from (i ),(ii) and (iii) we get=>

ST = PR - PQ

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

Hence proved