Question

If $sin\Theta =\frac{3}{4}$, Prove that $\frac{\sqrt{cosec^{2}\Theta-cot^{2}\Theta } }{sec^{2}\Theta-1 } =\frac{\sqrt{7} }{3}$

Answer 1

SOLUTION :  

Given: sin θ =¾

sin θ = ¾ = P/H = BC/AC

Perpendicular side = 3 and Hypotenuse = 4

In right angled ΔABC, by using Pythagoras theorem

AC² = AB² + BC²

4² = AB² + 3²  

16 = AB² + 9  

AB²  = 16 -  9

AB² = 7

AB² = √7

Base (AB) = √7  

Now, cosec θ = Hypotenuse/ Perpendicular

cosec θ = 4/3

cot θ = Base / perpendicular

cot θ = √7 / 4

sec θ = Hypotenuse / base

sec θ = 4/√7

CALCULATION of prove that is in the ATTACHMENT.

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