Question

14.
If 0 is a positive acute angle such that seco=cosec60°, find the value of 2cos' 0-1​

Answer 1

SOLUTION :  

Given : sec θ = cosec 60° and  θ is positive acute angle .

sec θ = cosec 60°

cosec (90 - θ) = cosec 60°

[cosec (90 - θ) = sec θ]

On equating both sides,

(90 - θ) = 60°

θ = 90° - 60°

θ = 30°

We have to find a value :  2cos²θ - 1  

2cos²θ - 1  

Put θ = 30°

2cos²θ - 1  = 2 × cos² 30° - 1

= 2 × (√3/2)² - 1  

[cos 30° = √3/2]

= 2 × ¾ - 1

= 3/2 - 1

=( 3 - 2)/2

= ½

Hence, the value of 2cos²θ - 1 is ½ .

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Answer 2

Step-by-step explanation:

Given : sec θ = cosec 60° and  θ is positive acute angle .

sec θ = cosec 60°

cosec (90 - θ) = cosec 60°

[cosec (90 - θ) = sec θ]

On equating both sides,

(90 - θ) = 60°

θ = 90° - 60°

θ = 30°

We have to find a value :  2cos²θ - 1  

2cos²θ - 1  

Put θ = 30°

2cos²θ - 1  = 2 × cos² 30° - 1

= 2 × (√3/2)² - 1  

[cos 30° = √3/2]

= 2 × ¾ - 1

= 3/2 - 1

=( 3 - 2)/2

= ½

Hence, the value of 2cos²θ - 1 is ½ .