Question

If cot theta = 1/(sqrt(3)) the value of sec^2 theta + cosec ^ 2 theta​

Answer 1

Step-by-step explanation:

Given:

Cot θ = √2 + 1.

Concept Used:

Cot θ = Base/Perpendicular

Cosec θ = Hypotenuse/Perpendicular

Sec θ = Hypotenuse/Base

Calculation:

Cot θ = B/P = (√2 + 1)/1

By Pythagoras theorem,

⇒ H2 = P2 + B2

⇒ H2 = 12 + (√2 + 1)2

⇒ H2 = 1 + 2 + 1 + 2√2

⇒ H2 = 4 + 2√2

Cosec θ × Sec θ

⇒ H/P × H/B

⇒ H2/BP

⇒ (4 + 2√2)/(√2 + 1) × 1

⇒ (4 + 2√2)/(√2 + 1)

⇒ [(√2 + 1) × (2√2)] / (√2 + 1)

⇒ 2√2

∴ The value of Cosec θ × Sec θ is 2√2.

We know,

Cosec2θ = 1 + Cot2θ

Sec2θ = 1 + Tan2θ

So, Sec θ × Cosec θ

⇒ √(1 + tan2θ) × √(1 + Cot2θ)

⇒ √ [1 + {1/(√2 + 1)}2] × [1 + (√2 + 1)2] ....(i)

The value of 1/(√2 + 1)

⇒ 1× (√2 - 1) /(√2 + 1) × (√2 - 1)

⇒ (√2 - 1)/2 - 1

⇒ (√2 - 1)

Putting in equation (i)

⇒ √ [1 + {1/(√2 + 1)}2] × [1 + (√2 + 1)2]

⇒ √ [1 + (√2 - 1)2] × [1 + (√2 + 1)2]

⇒ √(4 - 2√2) × (4 - 2√2)

⇒ √16 - 8 = √8

⇒ 2√2

∴ The value of Cosec θ × Sec θ is 2√2.