Lutasin ang tan^2θ-sec^2θ=

$$\tan ^ { 2 } \theta - \sec ^ { 2 } \theta =$$

I-factor

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Mga Hakbang ng Solution

Maaaring i-factor ang difference ng mga square gamit ang panuntunang: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(\tan(\theta )-\frac{1}{\cos(\theta )}\right)\left(\tan(\theta )+\frac{1}{\cos(\theta )}\right)$$

Isaalang-alang ang \tan(\theta )-\frac{1}{\cos(\theta )}. I-factor out ang \frac{1}{\cos(\theta )}.

$$\frac{\sin(\theta )-1}{\cos(\theta )}$$

Isaalang-alang ang \tan(\theta )+\frac{1}{\cos(\theta )}. I-factor out ang \frac{1}{\cos(\theta )}.

$$\frac{\sin(\theta )+1}{\cos(\theta )}$$

I-rewrite ang kumpletong naka-factor na expression.

$$\left(\sin(\theta )-1\right)\left(\sin(\theta )+1\right)\times \left(\frac{1}{\cos(\theta )}\right)^{2}$$

I-evaluate

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