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Question Number 197346 by Amidip last updated on 14/Sep/23

	Answered by som(math1967) last updated on 14/Sep/23

	$\frac{a + b}{a - b} = \frac{t a n (θ + ϕ)}{t a n (θ - ϕ)}$ $\frac{a + b}{a - b} = \frac{s i n (θ + ϕ) c o s (θ - ϕ)}{s i n (θ - ϕ) c o s (θ + ϕ)}$ $\frac{2 a}{2 b} = \frac{s i n 2 θ}{s i n 2 ϕ} [u s i n g c o m p o n e n d o & d i v i d e n d o]$ $a s i n 2 ϕ = b s i n 2 θ$ $a^{2} {s i n}^{2} 2 ϕ = b^{2} {s i n}^{2} 2 θ$ $a^{2} - b^{2} = a^{2} {c o s}^{2} 2 ϕ - b^{2} {c o s}^{2} 2 θ$ $n o w g i v e n$ $a c o s 2 ϕ + b c o s 2 θ = c$ ${(a c o s 2 ϕ - c)}^{2} = b^{2} {c o s}^{2} 2 θ$ $a^{2} {c o s}^{2} 2 ϕ - 2 a c \cos 2 ϕ + c^{2} = b^{2} \cos^{2} 2 θ$ $a^{2} {c o s}^{2} 2 ϕ - b^{2} {c o s}^{2} 2 θ + c^{2} = 2 a c c o s 2 ϕ$ $a^{2} - b^{2} + c^{2} = 2 a c c o s 2 ϕ$ $[∵ a^{2} {c o s}^{2} 2 ϕ - b^{2} {c o s}^{2} 2 θ = a^{2} - b^{2}]$
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