Question

If sin(A-B)=1/2 , cos (A + B)=1/2 , then Sin(A+B)​

Answer 1

$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

☆$\sf\implies sin ( A - B ) = \dfrac{1}{2}$

$\sf{\red{\boxed{\bold{sin 30 = \dfrac{1}{2} }}}}$

$\sf\implies sin ( A - B ) = sin 30$

$\sf\implies A -B = 30$

$\sf{\green{\boxed{\bold{A - B = 30 }}}}$-----( i )

☆$\sf\implies cos( A + B ) = \dfrac{1}{2}$

$\sf{\red{\boxed{\bold{cos 60 = \dfrac{1}{2} }}}}$

$\sf\implies sin ( A - B ) = cos 60$

$\sf\implies A -B = 60$

$\sf{\green{\boxed{\bold{A - B = 60 }}}}$------ ( ii )

• Adding eq i and ii

$\sf\dag A + B + A - B = 60 + 30$

$\sf\dag 2A = 90$

$\sf\dag A = 45$

$\sf{\green{\boxed{\bold{A = 45 }}}}$

• putting value of A in eq i

$\dag A + B = 60$

$\dag 45 + B = 60$

$\dag B = 60 - 45$

$\dag B = 15$

$\sf{\green{\boxed{\bold{B = 15 }}}}$

_________________________

$\sf\longrightarrow sin A + B$

$\sf\longrightarrow sin 45 + 15$

$\sf\longrightarrow sin 60$

$\sf\longrightarrow \dfrac{\sqrt 3}{2}$

$\sf{\green{\boxed{\bold{\dag tan ( A + B) =\dfrac{\sqrt 3}{2} }}}}$

Answer 2

Answer:

Solution

[tex]

☆\sf\implies sin ( A - B ) [\tex]

\sf{\red{\boxed{\bold{sin 30 = \dfrac{1}{2} }}}}

sin30=

2

1

\sf\implies sin ( A - B ) = sin 30⟹sin(A−B)=sin30

\sf\implies A -B = 30⟹A−B=30

\sf{\green{\boxed{\bold{A - B = 30 }}}}

A−B=30

-----( i )

☆\sf\implies cos( A + B ) = \dfrac{1}{2}⟹cos(A+B)=

2

1

\sf{\red{\boxed{\bold{cos 60 = \dfrac{1}{2} }}}}

cos60=

2

1

\sf\implies sin ( A - B ) = cos 60⟹sin(A−B)=cos60

\sf\implies A -B = 60⟹A−B=60

\sf{\green{\boxed{\bold{A - B = 60 }}}}

A−B=60

------ ( ii )

Adding eq i and ii

\sf\dag A + B + A - B = 60 + 30†A+B+A−B=60+30

\sf\dag 2A = 90†2A=90

\sf\dag A = 45†A=45

\sf{\green{\boxed{\bold{A = 45 }}}}

A=45

putting value of A in eq i

\dag A + B = 60†A+B=60

\dag 45 + B = 60†45+B=60

\dag B = 60 - 45†B=60−45

\dag B = 15†B=15

\sf{\green{\boxed{\bold{B = 15 }}}}

B=15

_________________________

\sf\longrightarrow sin A + B⟶sinA+B

\sf\longrightarrow sin 45 + 15⟶sin45+15

\sf\longrightarrow sin 60⟶sin60

\sf\longrightarrow \dfrac{\sqrt 3}{2}⟶

2

3

\sf{\green{\boxed{\bold{\dag tan ( A + B) =\dfrac{\sqrt 3}{2} }}}}

†tan(A+B)=

2

3

[\tex]