Question

If cos (90

-A)=0, then Sin (A+B)=​

Answer 1

Given :- cos(90 - A) = 0

To Find :- sin(A + B) = ?

Solution :-

we know that,

• cos(90 - x) = sin x .
• sin(x + y) = sin x * cos y + cos x * sin y

so,

→ cos(90 - A) = 0

→ sin A = 0 ---- Eqn.(1)

then,

→ sin(A + B)

→ sin A * cos B + cos A * sin B

putting value of Eqn.(1)

→ 0 * cos B + cos A * sin B

→ cos A * sin B ------ Eqn.(2)

now, since,

→ sin A = 0

→ sin A = sin 0°

→ A = 0°

therefore,

→ cos A * sin B

→ cos 0° * sin B

→ 1 * sin B

→ sin B (Ans.)

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Answer 2

Given :  Cos (90° - A)  = 0

To Find :  Sin (A + B)

Solution:

Method 1 :

Cos (90° - A)  = 0

=> Sin A = 0

=> A = nπ

Sin (A + B) = Sin ( nπ + B)  = ± SinB

Method 2 :

Cos (90° - A)  = 0

=> Sin A = 0

Sin²A + Cos²A = 1

=> Cos²A  = 1

=> Cos A = ±1

Sin (A + B) = SinACosB + CosASinB

=> Sin (A + B) =  0 *CosB +  ( ±1)SinB

=> Sin (A + B) =     0   ±  SinB

=> Sin (A + B) =    ±  SinB

Sin (A + B) =    ±  SinB

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