which equation can be used to prove ( 1+\tan ^{2}(x)=sec ^{2}(x) )?\n( \frac{cos ^{2}(x)}{sec…

which equation can be used to prove ( 1+\tan ^{2}(x)=sec ^{2}(x) )?\n( \frac{cos ^{2}(x)}{sec ^{2}(x)}+\frac{sin ^{2}(x)}{sec ^{2}(x)}=\frac{1}{sec ^{2}(x)} )\n( \frac{cos ^{2}(x)}{sin ^{2}(x)}+\frac{sin ^{2}(x)}{sin ^{2}(x)}=\frac{1}{sin ^{2}(x)} )\n( \frac{cos ^{2}(x)}{\tan ^{2}(x)}+\frac{sin ^{2}(x)}{\tan ^{2}(x)}=\frac{1}{\tan ^{2}(x)} )\n( \frac{cos ^{2}(x)}{cos ^{2}(x)}+\frac{sin ^{2}(x)}{cos ^{2}(x)}=\frac{1}{cos ^{2}(x)} )

which equation can be used to prove ( 1+\tan ^{2}(x)=sec ^{2}(x) )?\n( \frac{cos ^{2}(x)}{sec ^{2}(x)}+\frac{sin ^{2}(x)}{sec ^{2}(x)}=\frac{1}{sec ^{2}(x)} )\n( \frac{cos ^{2}(x)}{sin ^{2}(x)}+\frac{sin ^{2}(x)}{sin ^{2}(x)}=\frac{1}{sin ^{2}(x)} )\n( \frac{cos ^{2}(x)}{\tan ^{2}(x)}+\frac{sin ^{2}(x)}{\tan ^{2}(x)}=\frac{1}{\tan ^{2}(x)} )\n( \frac{cos ^{2}(x)}{cos ^{2}(x)}+\frac{sin ^{2}(x)}{cos ^{2}(x)}=\frac{1}{cos ^{2}(x)} )

Answer

Explanation:

Step1: Recall the fundamental trigonometric identity

We know that (\sin^{2}(x)+\cos^{2}(x) = 1).

Step2: Divide both sides of the identity (\sin^{2}(x)+\cos^{2}(x)=1) by (\cos^{2}(x))

Using the rule (\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}), we get (\frac{\sin^{2}(x)}{\cos^{2}(x)}+\frac{\cos^{2}(x)}{\cos^{2}(x)}=\frac{1}{\cos^{2}(x)}). Since (\tan(x)=\frac{\sin(x)}{\cos(x)}), then (\tan^{2}(x)=\frac{\sin^{2}(x)}{\cos^{2}(x)}), and (\sec(x)=\frac{1}{\cos(x)}), so (\sec^{2}(x)=\frac{1}{\cos^{2}(x)}).

Answer:

(\frac{\cos^{2}(x)}{\cos^{2}(x)}+\frac{\sin^{2}(x)}{\cos^{2}(x)}=\frac{1}{\cos^{2}(x)}) (the fourth option)