Question 51

If $$\tan \theta + \sec \theta = \frac{(x - 2)}{(x + 2)}$$, then what is the value of $$\cos \theta$$?

Solution

$$\tan\theta+\sec\theta=\frac{(x-2)}{(x+2)}$$

we know, $$\left(\tan\theta+\sec\theta\right)\left(\sec\theta\ -\tan\theta\ \right)=1\ .$$

So, $$\left(\sec\theta\ -\tan\theta\ \right)=\frac{x+2}{x-2}\ .$$

So, $$2\sec\theta\ =\frac{x+2}{x-2}\ +\frac{x-2}{x+2}=\frac{x^2+4x+4+x^2-4x+4}{x^2-4}=\frac{2\left(x^2+4\right)}{x^2-4}\ .$$

or, $$\cos\theta\ =\frac{x^2-4}{x^2+4}\ .$$

So, C is correct choice.

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