Question

If sin theta = 3/4 then prove that whole under root cosec square theta - cot square theta by sec square theta -1 is equal to under root 7 by 3

Answer 1

Answer:

$\sqrt{\frac{Cosec^2\theta - cot^2\theta}{Sec^2\theta - 1}}=\frac{\sqrt{7}}{3}$

Step-by-step explanation:

Sinθ = 3/4

Sin²θ = 9/16

Cos²θ = 1 - Sin²θ  = 7/16

$\sqrt{\frac{Cosec^2\theta - cot^2\theta}{Sec^2\theta - 1}}=\frac{\sqrt{7}}{3}$

Cosec²θ = 1/Sin²θ  = 1/(9/16) = 16/9

Cot²θ = Cos²θ/Sin²θ = (7/16)/(9/16) = 7/9

Cosec²θ - Cot²θ = 16/9 - 7/9 = 9/9 = 1

Sec²θ = 1/Cos²θ = 1/(7/16) = 16/7

Sec²θ - 1 = 16/7 - 1 = 9/7

putting all these values

LHS =

$\sqrt{\frac{1}{\frac{9}{7}}} \\=\sqrt{\frac{7}{9} } \\= \frac{\sqrt{7}}{3}$

= RHS

Answer 2

Answer:

Step-by-step explanation: