Question

1] If V1 + x2 sine = x, prove that tan’e + cote = x2 +1/x2​

Answer 1

Required Answer:-

Given:

• √(1 + x²) sin θ = x

To Prove:

• tan²θ + cot²θ = x² + 1/x²

Proof:

Let ABC be a right angled triangle right angled at B and ∠ACB = θ

Therefore,

➡ √(1 + x²) sin θ = x

➡ sin θ = x/√(1 + x²)

As sin θ is the ratio of perpendicular to the Hypotenuse, therefore,

➡ sin θ = Perpendicular/Hypotenuse.

Let Perpendicular = kx and Hypotenuse = k√(1 + x²) where k is the common ratio.

From fig,

➡ AC² = AB² + BC² (By Pythagoras Theorem)

➡ [k√(1 + x²)]² = (kx) ² + BC²

➡ BC² = k²(1 + x²) - k²x²

➡ BC² = k²(1 + x² - x²)

➡ BC = √(k² × 1)

➡ BC = 1k

Therefore,

➡ Base = 1k

➡ tan θ = Perpendicular/Base

➡ tan θ = kx/k

➡ tan θ = x

➡ cot θ = 1/x

Now, taking LHS,

tan²θ + cot²θ

= (x)² + (1/x)²

= x² + 1/x²

= RHS (Hence Proved)