If cos θ = 12/13, 0 < θ < π /2 find the value of ​​​(sin^2 θ - cos^2θ )/(2sin θ .sin θ), 1/tan^2 θ

Question

If cos θ = \(\frac{12}{13}\), 0 < θ < \(\frac{\pi}{2}\) find the value of 

\(\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta \cos\theta}, \frac{1} {\tan^2\theta}\)

(sin²θ - cos²θ )/(2sin θ .sin θ), 1/tan²θ 

Answer 1

cos θ = \(\frac{12}{13}\)

We know that

sin² θ = 1 – cos² θ

= 1 -(\(\frac {12}{13}\))²

= 1 - \(\frac {144}{169}\)

^{=\(\frac{25}{169}\)}

∴ sin θ = ± \(\frac{5}{13}\)

Since 0 < θ < \(\frac{\pi}{2}\) , θ lies in the 1st quadrant, 

∴ sin θ > 0

∴ sin θ = \(\frac{5}{13}\)