Question

please answer this question​

Answer 1

Solution :-

4sinθ = 3

⇒ sinθ = 3/4 = Perpendicular/Hypotenuse

Comparing on both sides

• Perpendicular = 3
• Hypotenuse = 4

By pythagoras theorem

⇒ (Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ 3² + (Base)² = 4²

⇒ 9 + (Base)² = 16

⇒ (Base)² = 16 - 9

⇒ Base = √7

$\sf \sqrt{ \dfrac{cosec^{2} \theta - cot^{2} \theta }{sec ^{2} \theta - 1 } } + 2cot \theta = \dfrac{ \sqrt{7} }{x} + cos \theta$

$\implies \sf \sqrt{ \dfrac{1 }{tan ^{2} \theta } } + 2cot \theta = \dfrac{ \sqrt{7} }{x} + cos \theta$

$\implies \sf \sqrt{ \bigg(\dfrac{1 }{tan \theta } \bigg)^{2} } + 2cot \theta = \dfrac{ \sqrt{7} }{x} + cos \theta$

$\implies \sf \dfrac{1 }{tan \theta } + 2cot \theta = \dfrac{ \sqrt{7} }{x} + cos \theta$

$\implies \sf cot \theta + 2cot \theta = \dfrac{ \sqrt{7} }{x} + cos \theta$

$\implies \sf 3cot \theta = \dfrac{ \sqrt{7} }{x} + cos \theta$

Required trignometric ratios :

• cosθ = Base/Hypotenuse = √7/4
• cot θ = Base/Perpendicular = (√7/4) / (3/4) = √7/3

Now substituting the values

$\implies \sf 3 \bigg( \dfrac{ \sqrt{7} }{3} \bigg) = \dfrac{ \sqrt{7} }{x} + \dfrac{ \sqrt{7} }{4}$

$\implies \sf \sqrt{7} = \dfrac{ \sqrt{7} }{x} + \dfrac{ \sqrt{7} }{4}$

$\implies \sf \sqrt{7} - \dfrac{ \sqrt{7} }{4} = \dfrac{ \sqrt{7} }{x}$

$\implies \sf \dfrac{4 \sqrt{7} - \sqrt{7} }{4} = \dfrac{ \sqrt{7} }{x}$

$\implies \sf \dfrac{3 \sqrt{7} }{4} = \dfrac{ \sqrt{7} }{x}$

$\implies \sf \dfrac{x}{4} = \dfrac{ \sqrt{7} }{ 3 \sqrt{7} }$

$\implies \sf \dfrac{x}{4} = \dfrac{1 }{ 3}$

$\implies \boxed{ \sf x = \dfrac{4 }{ 3} }$

Hence the value of x is 4/3.