# Trigonometry Important Question 14

#### Introduction to Trigonometry | Componendo Dividendo Rule

This class 10 Maths extra practice question is from chapter 8 - Introduction to 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 $$frac{\text{sinθ + 1}}{\text{sinθ − 1}}$ =$\frac{\sqrt{3} + 2}{\sqrt{3} - 2}$, find θ ## Target Centum in CBSE 10th Maths #### Online CBSE Course online.maxtute.com ### Video Explanation ## NCERT Solution to Class 10 Maths #### With Videos ### Explanatory Answer | Trigonometry Important Question 14 #### What is Componendo Dividendo Rule? If $\frac{\text{P}}{\text{Q}}$ = $\frac{\text{R}}{\text{S}}$ , then $\frac{\text{P + Q}}{\text{P - Q}}$= $\frac{\text{R + S}}{\text{R - S}}$ #### Apply Componendo Dividendo Rule $\frac{\text{sin θ + 1}}{\text{sin θ − 1}}$ =$\frac{\sqrt{3} + 2}{\sqrt{3} - 2}$ becomes $\frac{\text{$sin θ + 1$ + (sin θ − 1)}}{$text{$sin θ + 1) − (sin θ − 1)}}) = $$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°

#### Alternative Method

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°

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