Question 69

If $$\sec \theta = 8x and \tan \theta = \frac{8}{x}$$(x ≠ 0), then the value of $$16(x^2 - \frac{1}{x^2})$$ is:

Solution

As We know the trigonometry identity :

$$\sec ^2\theta - \tan^2\theta = 1$$

By putting value of $$\sec\theta$$ and $$\tan\theta$$, we get

$$\Rightarrow(8x)^2 - (\frac{8}{x})^2 = 1$$

$$\Rightarrow 64x^2 - \frac{64}{x^2} = 1$$

$$\Rightarrow 64(x^2 -\frac{1}{x^2}) = 1$$

$$\therefore 16(x^2 -\frac{1}{x^2}) = \frac{1}{4}$$

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