Question 75

If $$\tan^2 x — 3 \sec^2 x + 3 = 0,$$ then the value of $$x (0 \leq x \leq 90^\circ)$$ is:

Solution

$$\tan^2 x — 3 \sec^2 x + 3 = 0$$,

$$=$$> $$\tan^2x-3\left(1+\tan^2x\right)+3=0$$

$$=$$> $$\tan^2x-3-3\tan^2x+3=0$$

$$=$$> $$-2\tan^2x=0$$

$$=$$> $$\tan^2x=0$$

$$=$$> $$\tan x=0$$

$$=$$> $$x=0^{\circ\ }$$ since $$ (0 \leq x \leq 90^\circ)$$

Hence, the correct answer is Option B

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