If α and β are such that tan α = 2 tanβ, then what is sin (α + β

1.	If α and β are such that tan α = 2 tanβ, then what is sin (α + β) equal to?(a) 1 (b) 2 sin (α – β) (c) sin (α – β) (d) 3 sin (α – β)
Answer» (d) 3 sin (α - β) Given, tan α = 2 tan β ⇒ \(\frac{\frac{sin\,α}{cos\,α}}{\frac{sin\,β}{cos\,β}}=2\) = \(\frac{sin\,α\,cos\,β}{cos\,α\,sin\,β}=\frac21\) Using componendo and dividendo, we get \(\frac{sin\,α\,cos\,β+cos\,α\,sin\,β}{sin\,α\,cos\,β\,-cos\,α\,sin\,β}=\frac{2+1}{2-1}\) = \(\frac{sin\,(α+β)}{sin\,(α-β)}=\frac31\) ⇒ sin (α + β) = 3 sin (α - β).

If α and β are such that tan α = 2 tanβ, then what is sin (α + β) equal to?(a) 1 (b) 2 sin (α – β) (c) sin (α – β) (d) 3 sin (α – β)

Answer»

(d) 3 sin (α - β)

Given, tan α = 2 tan β ⇒ \(\frac{\frac{sin\,α}{cos\,α}}{\frac{sin\,β}{cos\,β}}=2\)

= \(\frac{sin\,α\,cos\,β}{cos\,α\,sin\,β}=\frac21\)

Using componendo and dividendo, we get

\(\frac{sin\,α\,cos\,β+cos\,α\,sin\,β}{sin\,α\,cos\,β\,-cos\,α\,sin\,β}=\frac{2+1}{2-1}\)

= \(\frac{sin\,(α+β)}{sin\,(α-β)}=\frac31\) ⇒ sin (α + β) = 3 sin (α - β).