Question

Find the value of [sec 3θ + tan 6θ + sin (70° + θ)]^-1, if sec(2θ - 10°) = cosec 3θ
1. 3 
2. √3 - 1
3. (3 + √3)/6
4. (3 - √2)/6

Answer 1

Correct Answer - Option 3 : (3 + √3)/6

Given :

sec(2θ - 10°) = cosec(3θ)

Formula used :

cosecθ = sec(90° - θ)

Calculations :

sec(2θ - 10°) = cosec(3θ)

sec(2θ - 10°) = sec(90° - 3θ)

Now 

⇒ 2θ - 10° = 90° - 3θ 

⇒ 5θ = 100° 

⇒ θ = 20° 

So, 

⇒ sec3θ + tan6θ + sin(70° + θ) = sec60° + tan120° + sin 90°  

⇒ 2 - √3 + 1          (tan120° = -cot30° = -√3)

⇒ 3 - √3 

[sec 3θ + tan 6θ + sin (70° + θ)]^-1 = (3 - √3)^-1

⇒ (3 + √3)/6

∴ Option 3 is the right choice.