Question

if cos (90°-A) =0,then sin (A+B)=​

Answer 1

Given:—

• cos (90⁰ - A) = 0

$→sinA \: = 0$

$→sinA = sin {0}$

$\underline \bold{→A = 0}$

Now,

$\bold{sin (A+B)}$

$= sin (0+B)$

$= \boxed{\bold{sin B}}$

♦Identity used :—

$\large{ \boxed{ \mathtt {\bold{cos (90⁰ - \theta) = sin\theta}}}}$

Answer 2

Answer:

sinB

Step-by-step explanation:

Given ,

cos(90° - A) = 0

To Find :-

Value of :-

sin(A + B)

How To Do :-

As they gave that 'cos(90° - A) = 0' we need to expand it by using the formula of 'cos(A - B)' and we need find the value of 'A' or by using the quadrants we can find the value of 'A' and we need to substitute the value of 'A' in sin(A + B).

Formula Required :-

cos(90° - α) = sinα

cos(A - B) = cosAcosB + sinAsinB

sin(A + B) = sinAcosB + cosAsinB

sin0° = 0

cos0° = 0

Solution :-

Method 1 :-

cos(90° - A) = 0

cos90°.cosA + sin90°.sinA = 0

(0) cosA + (1)sinA = 0

0 + sinA = 0

sinA = 0

sinA = sin0°

cancelling 'sin' on both sides :-

A = 0°

sin(A + B) = sinAcosB + cosAsinB

= sin0°.cosB + cos0°. sinB

= (0) cosB + (1) sinB

= 0 + sinB

= sinB

∴ sin(A + B) = sinB

Method 2 :-

cos(90° - A) = 0

→ sinA = 0

[ ∴ cos(90° - α) = sinα ]

sinA = sin0°

cancelling 'sin' on both sides :-

A = 0°

sin(A + B) = sin(0° + B)

= sinB

∴ sin(A + B) = sinB.