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Range of Trigonometric Ratios from 0 to 90 Degrees

if a2-b2sin...

Question

if $(a^{2} - b^{2}) sin θ + 2 a b cos θ = a^{2} + b^{2}$ , then prove $tan θ = \frac{a^{2} - b^{2}}{2 a b}$

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Solution

According to question.......
$\begin{matrix} (a^{2} - b^{2}) sin θ + 2 a b cos θ = a^{2} + b^{2} \Rightarrow (a^{2} + b^{2}) - (a^{2} - b^{2}) sin θ = 2 a b cos θ N o w, s q u a r i n g b o t h s i d e, \Rightarrow {((a^{2} + b^{2}) - (a^{2} - b^{2}) sin θ)}^{2} = (2 a b cos θ)^{2} \Rightarrow (a^{2} + b^{2})^{2} + (a^{2} - b^{2})^{2} {sin}^{2} θ - 2 (a^{2} + b^{2}) (a^{2} - b^{2}) sin θ = 4 a^{2} b^{^{2}} {cos}^{2} θ \Rightarrow (a^{2} + b^{2})^{2} + (a^{2} - b^{2})^{2} {sin}^{2} θ - 2 (a^{2} + b^{2}) (a^{2} - b^{2}) sin θ - 4 a^{2} b^{^{2}} (1 - s i n^{2} θ) = 0 \Rightarrow (a^{2} + b^{2})^{2} + (a^{2} - b^{2})^{2} {sin}^{2} θ - 2 (a^{2} + b^{2}) (a^{2} - b^{2}) sin θ - 4 a^{2} b^{2} + 4 a^{2} b^{2} {sin}^{2} θ) = 0 \Rightarrow [{(a^{2} - b^{2})}^{2} + 4 a^{2} b^{2}] {sin}^{2} θ - 2 (a^{2} + b^{2})^{2} (a^{2} - b^{2}) sin θ + [{(a^{2} + b^{2})}^{2} - 4 a^{2} b^{2}] = 0 \Rightarrow (a^{2} + b^{2})^{2} {sin}^{2} θ - 2 (a^{2} + b^{2})^{2} (a^{2} - b^{2}) sin θ + (a^{2} - b^{2})^{2} = 0 \Rightarrow {[{(a^{2} + b^{2})}^{2} sin θ - (a^{2} - b^{2})]}^{2} = 0 \Rightarrow (a^{2} + b^{2}) sin θ - (a^{2} - b^{2}) = 0 \Rightarrow sin θ = \frac{(a^{2} - b^{2})}{(a^{2} + b^{2})} \Rightarrow cos e c θ = \frac{a^{2} + b^{2}}{a^{2} - b^{2}} N o w, {cot}^{2} θ = cos e c^{2} θ - 1 \Rightarrow {cot}^{2} θ = {(\frac{a^{2} + b^{2}}{a^{2} - b^{2}})}^{2} - 1 \Rightarrow {cot}^{2} θ = \frac{{(a^{2} + b^{2})}^{2} - {(a^{2} - b^{2})}^{2}}{{(a^{2} - b^{2})}^{2}} = \frac{(a^{2} + b^{2} - a^{2} + b^{2}) (a^{2} + b^{2} + a^{2} - b^{2})}{{(a^{2} - b^{2})}^{2}} \Rightarrow {cot}^{2} θ = \frac{(2 a^{2}) (2 b^{2})}{{(a^{2} - b^{2})}^{2}} = \frac{4 a^{2} b^{2}}{{(a^{2} - b^{2})}^{2}} \Rightarrow cot θ = \sqrt{\frac{4 a^{2} b^{2}}{{(a^{2} - b^{2})}^{2}}} = \frac{2 a b}{a^{2} + b^{2}} \Rightarrow tan θ = \frac{a^{2} + b^{2}}{2 a b} ∣ ∣ cot θ = \frac{1}{tan θ} s o t h a t w e p r o v e : tan θ = \frac{a^{2} + b^{2}}{2 a b} \end{matrix}$

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Similar questions

(a^{2} - b^{2}) sin θ + 2 a b cos θ = a^{2} + b^{2}

tan θ

(a^{2} - b^{2}) sin θ + 2 a b cos θ = a^{2} + b^{2}

then find the value of

tan θ

sin α = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}

then prove that

cot α = \frac{2 a b}{a^{2} - b^{2}}

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(c) c² + a² − b² + 2ca = 4s (s − b)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

I f (a^{2} - b^{2}) sin θ + 2 a b cos θ = a^{2} + b^{2} f i n d tan θ

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