Question

If sin????=3/4, prove that √cosec²????-cot²????/sec²????-1=√7/3

Answer 1

Answer:

please write proper question and please follow me

Answer 2

√cosec²θ-cot²θ / sec²θ-1 = √7/3  proved

Step-by-step explanation:

Given: sinθ = 3/4

To find: √cosec²θ-cot²θ / sec²θ-1 = √7/3

Solution:  

Sinθ = Opposite / Hypotenuse = 3/4

Adjacent  =  √Hypotenuse² - Opposite² = √(4²- 3²) = √16-9 = √7

Cosθ = Adjacent / Hypotenuse =  √7/4

 Cosecθ = 1/sinθ = 4/3

 Secθ = 1/cosθ = 4/√7

Cotθ = Cosθ/sinθ = √7/3

Now LHS = √cosec²θ-cot²θ / sec²θ-1  

  = √ (4/3)² - (√7/3)² / (4/√7)² - 1

  = √ (16/9 - 7/9) / 16/7 -1

  = √ 9/9 / (9/7)

  = √ 1 / (9/7)

  = √7/9

 = √7/3

  = RHS

  LHS = RHS. Hence proved.