Q.39. If $$ \sec heta=4 x $$ and $$ an heta=\frac{4}{x},(x
eq 0) $$ then the value of $$ 8\left(x^{2}-\frac{1}{x^{2}} ight) $$ is : SSC CHSL $$ 05 / 07 / 2019 $$ (Afternoon)

Question

Q.39. If $$ \sec 	heta=4 x $$ and $$ 	an 	heta=\frac{4}{x},(x 
eq 0) $$ then the value of $$ 8\left(x^{2}-\frac{1}{x^{2}}ight) $$ is : SSC CHSL $$ 05 / 07 / 2019 $$ (Afternoon)

UpStudy AI · Accepted Answer

To solve the problem, let's start by using the trigonometric identity:

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

Given:
$$
\sec \theta = 4x \quad \text{and} \quad \tan \theta = \frac{4}{x}
$$

Square both sides of the given equations:
$$
\sec^2 \theta = (4x)^2 = 16x^2
$$
$$
\tan^2 \theta = \left(\frac{4}{x}\right)^2 = \frac{16}{x^2}
$$

Substitute these into the identity:
$$
16x^2 - \frac{16}{x^2} = 1
$$

Factor out the 16:
$$
16\left(x^2 - \frac{1}{x^2}\right) = 1
$$

Divide both sides by 16:
$$
x^2 - \frac{1}{x^2} = \frac{1}{16}
$$

Now, multiply both sides by 8 to find the desired expression:
$$
8\left(x^2 - \frac{1}{x^2}\right) = 8 \times \frac{1}{16} = \frac{1}{2}
$$

**Answer:** The value of $$ 8\left(x^{2}-\frac{1}{x^{2}}\right) $$ is $$\frac{1}{2}$$.
The value of $$ 8\left(x^{2}-\frac{1}{x^{2}}\right) $$ is $$\frac{1}{2}$$.

Q.39. If and
then the value of is :
SSC CHSL (Afternoon)

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Q.39. If and then the value of is : SSC CHSL (Afternoon)