Question

if cos(a+ß)=0,then sin(a+ß) can be reduced to-
a. cos ß
b. cos2ß
c. sin a
d. sin2a​

Answer 1

Answer:

(b) Cos2β

Step-by-step explanation:

I think question is like this Cos ( α - β ) = 0 then find value of Sin ( α + β )

Given---> Cos( α - β ) = 0

To find ---> Sin ( α + β ) = ?

Solution---> ATQ,

Cos ( α - β ) = 0

=> Cos ( α - β ) = Cos 90°

=> α - β = 90°

=> α = 90° + β

Now , Sin(α + β ) = Sin { ( 90° + β ) + β }

= Sin ( 90° + β + β )

= Sin ( 90° + 2β )

We know that Sin ( 90° + θ ) = Cοsθ , using it here, we get,

= Cos 2β

Additional information---->

1) Cos ( 90° +A ) = - SinA

2) tan ( 90° + A ) = - CotA

3) Cot ( 90° + A ) = - tanA

4) Sec ( 90° + A ) = - CosecA

5) Cosec ( 90° + A ) = SecA

Answer 2

Good Afternoon !

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i thnk ur question is like this↓↓↓↓

Question;

if cos(a+ß)=0,then sin(a-ß) can be reduced to-

a. cos ß

b. cos2ß

c. sin a

d. sin2a​

Step-by-step explanation:

Given, cos(α+β)=0=cos90∘    [∴cos90∘=0]

⇒α+β=90∘

⇒α=90∘−β……………..(i)

Now, sin(α−β)=sin(90∘−β−β) [put the value from Eq.(i)]

=sin(90∘−2β)

=cos2β     [∴sin(90∘−θ)=cosθ]

Hence, sin(α−β) can be reduced to cos2β.

Then Option B is Correct

if cos(a-ß)=0,then sin(a+ß) can be reduced to-