if sin(A+B)=1 and cos(A-B)=1;0°<A+B<=90°,A>=B,find A and B
Answers
EXPLANATION.
⇒ sin(A + B) = 1. - - - - - (1).
⇒ cos(A - B) = 1. - - - - - (2).
⇒ 0° < A + B ≤ 90°.
⇒ A > B.
As we know that,
We can write equation as,
⇒ sin(A + B) = sin(90°). - - - - - (1).
⇒ cos(A - B) = cos(0°). - - - - - (2).
We get,
⇒ A + B = 90°. - - - - - (1).
⇒ A - B = 0°. - - - - - (2).
Adding equation (1) & (2), we get.
⇒ 2A = 90°.
⇒ A = 45°.
Put the value of A = 45° in equation (1), we get.
⇒ A + B = 90°.
⇒ 45° + B = 90°.
⇒ B = 90° - 45°.
⇒ B = 45°.
Value of A = 45° & B = 45°.
MORE INFORMATION.
(1) = sin²θ + cos²θ = 1.
(2) = 1 + tan²θ = sec²θ.
(3) = 1 + cot²θ = cosec²θ.
❍ Given that, If sin( A + B ) = 1 and cos( A - B ) = 1 , 0° < A+B < = 90°, A>= B and we need to find A and B.
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Information given ::
- sin(A + B) = 1
- cos(A - B) = 1
- 0° < A + B ≤ 90°
- A > B
As we know that,
We can write the equations as,
⇒ sin( A + B ) = sin( 90° )
⇒ cos( A - B ) = cos( 0° )
H E N C E
⇒ A + B = 90° - [ i ]
⇒ A - B = 0° - [ ii ]
- Adding equation [ i ] and [ ii ]
➵ 2A = 90°
➵ A = 90°/2
➵ A = 45°
- Putting the value of A = 45° in equation [ i ]
➵ A + B = 90°
➵ 45° + B = 90°
➵ B = 90° - 45°
➵B = 45°
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Henceforth, value of A = 45° and B = 45°