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Trigonometric Ratios of Compound Angles

If cos α+β si...

Question

If $c o s (α + β) s i n (γ + δ) = c o s (α - β) s i n (γ - δ), p r o v e t h a t c o t α c o t β c o t γ = c o t δ$

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Solution

We have,
$c o s (α + β) s i n (γ + δ) = c o s (α - β) s i n (γ - δ) \Rightarrow \frac{c o s (α + β)}{c o s (α - β)} = \frac{s i n (γ - δ)}{s i n (γ + δ)}$
Now,

$\frac{c o s (α + β)}{c o s (α - β)} = \frac{s i n (γ - δ)}{s i n (γ + δ)} \Rightarrow \frac{c o s (α + β)}{c o s (α - β)} + 1 = \frac{s i n (γ - δ)}{s i n (γ + δ)} + 1 \Rightarrow \frac{c o s (α + β) + c o s (α - β)}{c o s (α - β)} = \frac{s i n (γ - δ) + s i n (γ + δ)}{s i n (γ + δ)} \frac{c o s (α + β)}{c o s (α - β)} = \frac{s i n (γ - δ)}{s i n (γ + δ)}$
[By equation(i)]
$\Rightarrow \frac{c o s (α + β)}{c o s (α - β)} - 1 = \frac{s i n (γ - δ)}{s i n (γ + δ)} - 1$

$\Rightarrow \frac{c o s (α + β) - c o s (α - β)}{c o s (α - β)} = \frac{s i n (γ - δ) - s i n γ + δ)}{s i n (γ + δ)}$
Dividing equation (ii) by equation (iii),

we get
$\frac{c o s (α + β) - c o s (α - β)}{c o s (α + β) - c o s (α - β)} = \frac{s i n (γ - δ) + s i n γ + δ)}{s i n (γ - δ) - s i n (γ - δ)}$

$\Rightarrow \frac{c o s (α + β) + c o s (α - β)}{c o s (α + β) - c o s (α - β)} = - [\frac{s i n (γ + δ) + s i n (γ - δ)}{s i n (γ + δ) - s i n (γ - δ)}]$

$= \frac{2 c o s {\frac{α + β + α - β}{2}} c o s {\frac{α + β - α + β}{2}}}{- 2 s i n {\frac{α + β + α - β}{2}} s i n {\frac{α + β - α + β}{2}}} ⎡ ⎢ ⎣ \frac{2 s i n {\frac{γ + δ + γ - δ}{2}} c o s {\frac{γ + δ - γ + δ}{2}}}{2 s i n - {\frac{γ + δ - γ + δ}{2}} c o s {\frac{γ + δ + γ - δ}{2}}} ⎤ ⎥ ⎦ \Rightarrow \frac{c o s α c o s β}{s i n α s i n β} = \frac{s i n γ c o s δ}{s i n δ c o s γ} \Rightarrow c o t α c o t β = \frac{s i n γ c o s δ}{c o s γ s i n δ} \Rightarrow c o t α c o t β = \frac{c o t δ}{c o t γ}$

$c o t α c o t β c o t γ = c o t δ$
$∴ c o t α c o t β c o t γ = c o t δ$

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