Question

Solve 7b plz
(Transformation of trigonometric formulae )
#salute u if u solved it

Answer 1

Solution :-

Given to prove :-

$\small{\dfrac{Cos(5\alpha)Sin(2\alpha) - Cos(4\alpha)Sin(3\alpha)}{Sin(5\alpha)Sin(2\alpha) - Cos(4\alpha)Cos(3\alpha)} = Cot(2\alpha)}$

Now multiply both numerator and denominator by 2 .

$\small{\rightarrow \dfrac{2.Cos(5\alpha)Sin(2\alpha) - 2.Cos(4\alpha)Sin(3\alpha)}{2.Sin(5\alpha)Sin(2\alpha) - 2.Cos(4\alpha)Cos(3\alpha)}}$

Now as

Sin(a + b) - Sin(a - b) = 2.Cos(a).Sin(b)

Cos(a - b) - Cos(a + b) = 2.Sin(a).Sin(b)

Cos(a - b) + Cos(a + b) = 2.Cos(a).Cos(b)

So our expression became :-

$\small{\rightarrow \dfrac{[Sin( 5\alpha + 2\alpha) - Sin( 5\alpha - 2\alpha)] - [ Sin( 4\alpha + 3\alpha) - Sin( 4\alpha - 3\alpha)] }{[Cos( 5\alpha - 2\alpha) - Cos( 5\alpha + 2\alpha)] - [ Cos( 4\alpha + 3\alpha) + Cos( 4\alpha - 3\alpha)] }}$

$\small{ \rightarrow \dfrac{[Sin( 7\alpha) - Sin( 3\alpha)] - [ Sin( 7\alpha ) - Sin(\alpha)] }{[Cos( 3\alpha) - Cos( 7\alpha)] - [ Cos( 7\alpha) + Cos(\alpha)]}}$

$\rightarrow \dfrac{Sin(\alpha) - Sin( 3\alpha) }{Cos( 3\alpha) - Cos(\alpha)}$

Now as we have

$Sin(C) - Sin(D) = 2.Cos \left(\dfrac{C+D}{2}\right)Cos \left(\dfrac{C-D}{2}\right)$

$Cos(C) - Cos(D) = -2.Sin \left(\dfrac{C+D}{2}\right)Cos \left(\dfrac{C-D}{2}\right)$

By applying it :-

$\small{\rightarrow \dfrac{2.Cos \left(\dfrac{\alpha + 3\alpha}{2}\right)Cos \left(\dfrac{\alpha - 3\alpha}{2}\right) }{-2.Sin\left(\dfrac{3\alpha + \alpha}{2}\right)Cos \left(\dfrac{3\alpha - \alpha}{2}\right)}}$

$\small{\rightarrow \dfrac{2.Cos \left(\dfrac{4\alpha}{2}\right)Cos \left(\dfrac{-2\alpha}{2}\right)}{2.Sin\left(\dfrac{4\alpha}{2}\right)Cos\left(\dfrac{2\alpha}{2}\right)}}$

$\rightarrow \dfrac{-2.Cos(2\alpha).Cos(\alpha)}{-2.Sin(2\alpha).Cos(\alpha)}$

$\rightarrow \dfrac{Cos(2\alpha)}{Sin(2\alpha)}$

$\rightarrow Cot(2\alpha)$

Now

$\rightarrow Cot(2\alpha) = Cot(2\alpha)$

As LHS = RHS

Hence Proved