If $$\sec \theta = 8x and \tan \theta = \frac{8}{x}$$(x ≠ 0), then the value of $$16(x^2 - \frac{1}{x^2})$$ is:
As We know the trigonometry identity :
$$\sec ^2\theta - \tan^2\theta = 1$$
By putting value of $$\sec\theta$$ and $$\tan\theta$$, we get
$$\Rightarrow(8x)^2 - (\frac{8}{x})^2 = 1$$
$$\Rightarrow 64x^2 - \frac{64}{x^2} = 1$$
$$\Rightarrow 64(x^2 -\frac{1}{x^2}) = 1$$
$$\therefore 16(x^2 -\frac{1}{x^2}) = \frac{1}{4}$$
Create a FREE account and get:
