This class 10 Maths extra practice question is from the topic Trigonometry. Concept Covered: Computing the value of angle by solving a trigonometric expression using Componendo Dividendo Rule. Level of difficulty: Hard

Question 14: If θ is an acute angle and \\frac{\text{sinθ + 1}}{\text{sinθ − 1}}) =\\frac{\sqrt{3} + 2}{\sqrt{3} - 2}), find θ

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If \\frac{\text{P}}{\text{Q}}) = \\frac{\text{R}}{\text{S}}) , then \\frac{\text{P + Q}}{\text{P - Q}})= \\frac{\text{R + S}}{\text{R - S}})

\\frac{\text{sin θ + 1}}{\text{sin θ − 1}}) =\\frac{\sqrt{3} + 2}{\sqrt{3} - 2}) becomes \\frac{\text{(sin θ + 1) + (sin θ − 1)}}{\text{(sin θ + 1) − (sin θ − 1)}}) = \\frac{(\sqrt{3}+ 2) + (\sqrt{3} − 2)}{(\sqrt{3}+ 2) - (\sqrt{3} − 2)})

\\frac{\text{2 sin θ }}{\text{2}})= \\frac{2 × \sqrt{3}}{\text{4}})

sin θ = \\frac{\sqrt{3}}{\text{2}})

We know that, sin 60° = \\frac{\sqrt{3}}{\text{2}})

∴ sin θ = sin 60°

**θ = 60°**

Divide both numerator and denominator of RHS by 2: \\frac{\frac{\sqrt{3} + 2}{2}}{\frac{\sqrt{3} - 2}{2}}) = \\frac{\frac{\sqrt{3}}{2} + 1 }{\frac{\sqrt{3}}{2} - 1 })

sin 60° = \\frac{\sqrt{3}}{2})

∴ \\frac{\frac{\sqrt{3}}{2} + 1 }{\frac{\sqrt{3}}{2} - 1 }) =\\frac{\text{sin 60° + 1}}{\text{sin 60° − 1}})

Compare the above result with LHS

\\frac{\text{ sin θ + 1 }}{\text{sin θ -1 }}) = \\frac{\text{ sin 60 ° + 1}}{\text{sin 60 ° − 1}})

∴ sin θ = sin 60°

**θ = 60°**

Class 10 Maths

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CBSE Class 10 Maths - 2020

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