Question

Somebody say answer for this.
This is 9th grade TRIGONOMETRY lesson.
Definitely i will make u as brainliest.​

Answer 1

Step-by-step explanation:

Given :-

Sin θ = a/√(a^2+b^2)

To show:-

b sin θ = a cos θ

Solution:-

Method -1:-

See the attachment

Method-2:-

Given that

Sin θ = a /√(a^2+b^2)

On squaring both sides then

=>Sin^2 θ = (a/√(a^2+b^2)^2

=>Sin^2 θ = a^2/(a^2+b^2)

=>1 - Sin^2 θ = 1-[a^2/(a^2+b^2)]

=>1-Sin^2 θ =[ (a^2+b^2)-a^2]/(a^2+b^2)

=>1 - Sin^2 θ = [a^2+b^2-a^2)/(a^2+b^2)

=>1-Sin^2 θ = b^2/(a^2+b^2)

We know that

Sin^2θ + Cos^2 θ =1

=>Cos^2 θ = b^2/(a^2+b^2)

=>Cos θ = √[b^2/(a^2+b^2)]

=>Cos θ = b/√(a^2+b^2)

Now

Cos θ/ Sin θ

=>[b/√(a^2+b^2)]/[a/√(a^2+b^2)]

=>[b/√(a^2+b^2)]×[√(a^2+b^2)/a]

On cancelling √(a^2+b^2) then

=>b/a

=>Cos θ/ Sin θ = b/a

=>a Cos θ = b Sin θ

Therefore , b Sin θ = a Cos θ

Hence, Proved

Used formula:-

• Sin^2θ + Cos^2 θ = 1