Question 65

If $$\sec \theta = 4x, and \tan \theta = \frac{4}{x}$$, (x ≠ 0) then the value of $$8\left(x^2 - \frac{1}{x^2}\right)$$ 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

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

$$\Rightarrow 16x^2 - \frac{16}{x^2} = 1$$

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

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

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