Question

if tan β = a/b , then cos β =?​

Answer 1

Answer:

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we have given,

atanα+btanβ=(a+b)tan(

2

α+β

)

⇒tanα+btanβ=atan(

2

α+β

)+btan(

2

α+β

)

⇒a[tanα−tan(

2

α+β

)]=b[tan(

2

α+β

)−tanβ]

usingtheformula:

tanA−tanB=

cosA.cosB

sin(A−B)

Now,substitutingtheformulabothside:

⇒a

cosα.cos(

2

α+β

)

sin(α−

2

α+β

)

=b

cosβ.cos(

2

α+β

)

sin(

2

α+β

−β)

⇒a

cosα

sin(α−

2

α+β

)

=b

cosβ

sin(

2

α+β

−β)

⇒a

cosα

sin(

2

α−β

)

=b

cosβ

sin(

2

α−β

)

⇒

cosα

a

=

cosβ

b

∴acosβ=bcosαprove.

Answer 2

Answer: cos β=$\frac{b}{\sqrt{a^{2} +b^{2} }}$

Step-by-step explanation:

tan β= a/b

or, $\frac{perpendicular}{base}=\frac{a}{b}$

now, hypotenuse will be=$\sqrt{a^{2} +b^{2} }$

We know that cos β=$\frac{base}{hypotenuse}$=$\frac{b}{\sqrt{a^{2} +b^{2} }}$

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