Question

If sin(A+B)=1 and cos(A-B)=1, 0°≤ (A+B) ≤ 90° and A=B, then find A and B.

Answer 1

Answer:

Sin(A+B)=Sin 90. (Sin 90=1)..1

Cos(A-B)=Cos 0. (Cos 0=1)..2

By 1 and 2 ,

we get

A+B=90

A-B=0

By solving both equation

we get answer as

A=45

AND,

B=45

Hope it might help you..

Thank you..

Answer 2

Answer:

Given,

sin(A+B)=1 ........(1)

cos(A-B)=1 ........(2)

We know that sin 90°=1 and cos0°=1

Therefore on applying the value of sin 90° and cos0° in equation (1) and (2)

We have,

sin(A+B)=1

A+B=90° .......(3)

Also,

cos(A-B)=1

A-B=0° .........(4)

On adding eq (3) and (4),

A+B=90°

A-B=0°

2A=90°

A=90°/2

=45°

Substitute the value of A in either equation (4)or(3)

Then, A+B=90°

45°+B=90°

B=90°-45°

=45°

therefore ,A=45° and B=45°

Thankyou hope this would help