If msin- n sin -2 - provethattan - - m - n - m - n

Given that m sinθ =n sin(θ+2α ) ∴sin(θ+2α )/sinθ =m/n Using componendo and dividendo, we get sin(θ+2α )+sinθ/sin(θ+2α ) sinθ=m+n/m n ⇒2sin ( θ+2α +θ/2 )cos ( θ+ ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Complex Numbers

If msinθ= n s...

Question

If $m s i n θ = n s i n (θ + 2 α), p r o v e t h a t t a n (θ + α) c o t α = \frac{m + n}{m - n}$

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Solution

Given that $m s i n θ = n s i n (θ + 2 α) ∴ \frac{s i n (θ + 2 α)}{s i n θ} = \frac{m}{n}$
Using componendo and dividendo, we get
$\frac{s i n (θ + 2 α) + s i n θ}{s i n (θ + 2 α) - s i n θ} = \frac{m + n}{m - n} \Rightarrow \frac{2 s i n (\frac{θ + 2 α + θ}{2}) c o s (\frac{θ + 2 α - θ}{2})}{2 c o s (\frac{θ + 2 α + θ}{2}) s i n (\frac{θ + 2 α - θ}{2})} = \frac{m + n}{m - n} [∵ s i n x + s i n y = 2 s i n \frac{x + y}{2} c o s \frac{x - y}{2} a n d s i n x - s i n y = 2 c o s \frac{x + y}{2} s i n \frac{x - y}{2}] \Rightarrow \frac{s i n (θ + α) c o s α}{c o s (θ + α) s i n α} = \frac{m + n}{m - n} \Rightarrow t a n (θ + α) c o t α = \frac{m + n}{m - n}$

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If msinθ=n sin(θ+2α),provethattan(θ+α)cotα=m+nm−n

If $m s i n θ = n s i n (θ + 2 α), p r o v e t h a t t a n (θ + α) c o t α = \frac{m + n}{m - n}$