Question 75

If $$\sec \theta = 3 x and \tan \theta = \frac{3}{x}$$, (x ≠ 0)then the value of $$9\left(x^2 - \frac{1}{x^2}\right)$$ is:

Solution

Given, $$\sec\theta=3x$$ and $$\tan\theta=\frac{\ 3}{x}$$

We know that, $$\sec^2\theta-\tan^2\theta=1$$

$$=$$> $$\left(3x\right)^2-\left(\frac{\ 3}{x}\right)^2=1$$

$$=$$> $$9x^2-\frac{\ 9}{x^2}=1$$

$$=$$> $$9\left(x^2-\frac{\ 1}{x^2}\right)=1$$

Hence, the correct answer is Option D

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