If sinθ = 12/13, find the value of sin^2θ - cos^2θ/2sinθ cosθ x 1/tan^2θ

Question

If sinθ = \(\frac{12}{13}\), find the value of \(\frac{sin^2θ-cos^2θ}{2sinθ\,cosθ}\times\frac{1}{tan^2θ}\)

Answer 1

Given: sinθ = \(\frac{12}{13}\)

To find: the value of \(\frac{sin^2θ-cos^2θ}{2sinθ\,cosθ}\times\frac{1}{tan^2θ}\)

Solution: Since sinθ = \(\frac{perpendicular}{hypotenuse}\)

So sinθ = \(\frac{12}{13}\) implies;

Perpendicular = AC = 12, Hypotenuse = BC = 13

Draw a right angled triangle at C,

By Pythagoras theorem,AB² = AC² + BC² 

⇒ (13)² = (12)² + BC² 

⇒ BC² = (13)² - (12)² 

⇒ BC² = 169 - 144 

⇒ BC² = 25

⇒ BC = \(\sqrt{25}\)

⇒ BC = 5

Since cosθ = Base/Hypotenuse and tanθ = Perpendicular/Base

⇒ cosθ = \(\frac{5}{13}\) and tanθ = \(\frac{12}{5}\)

Substitute the known values in \(\frac{sin^2θ-cos^2θ}{2sin^2θ\,cosθ}\) = \(\frac{1}{tan^2​​θ\,a}\),

⇒ \(\frac{sin^2θ-cos^2θ}{2sin^2θ\,cosθ}\) x \(\frac{1}{tan^2​​θ\,a}\)