Question 23

If $$2\sec^2 x - \tan^2 x = 5 and 0^\circ \leq x \leq 90^\circ$$ then $$x=?$$

Solution

$$2\sec^2 x - \tan^2 x = 5$$

$$\sec^2x+\sec^2x-\tan^2x=5$$

$$1+\sec^2x=5$$ ($$\sec^2x-\tan^2x=1$$)

$$\sec^2x=4$$

$$^{\sec^260^{\circ\ }=4\ \left(\sec60^{\circ\ }=2\right)}$$

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