Question
If $\cos\text{A}=\frac{7}{25},$ find is the value of $\tan\text{A}+\cot\text{A}$.
✓
Answer
$\cos\text{A}=\frac{7}{25}$
$\sin\text{A}=\sqrt{1-\cos^2\text{A}}=\sqrt{1-\Big(\frac{7}{25}\Big)^2}$
$=\sqrt{1-\frac{49}{625}}=\sqrt{\frac{625-49}{625}}$
$=\sqrt{\frac{576}{625}}=\frac{24}{25}$
Now, $\tan\text{A}+\cot\text{A}=\frac{\sin\text{A}}{\cos\text{A}}+\frac{\cos\text{A}}{\sin\text{A}}$
$=\frac{\frac{24}{25}}{\frac{7}{25}}+\frac{\frac{7}{25}}{\frac{24}{25}}=\frac{24}{25}\times\frac{25}{7}+\frac{7}{25}\times\frac{25}{24}$
$=\frac{24}{7}+\frac{7}{24}$
$=\frac{576+49}{168}=\frac{625}{168}$
$\sin\text{A}=\sqrt{1-\cos^2\text{A}}=\sqrt{1-\Big(\frac{7}{25}\Big)^2}$
$=\sqrt{1-\frac{49}{625}}=\sqrt{\frac{625-49}{625}}$
$=\sqrt{\frac{576}{625}}=\frac{24}{25}$
Now, $\tan\text{A}+\cot\text{A}=\frac{\sin\text{A}}{\cos\text{A}}+\frac{\cos\text{A}}{\sin\text{A}}$
$=\frac{\frac{24}{25}}{\frac{7}{25}}+\frac{\frac{7}{25}}{\frac{24}{25}}=\frac{24}{25}\times\frac{25}{7}+\frac{7}{25}\times\frac{25}{24}$
$=\frac{24}{7}+\frac{7}{24}$
$=\frac{576+49}{168}=\frac{625}{168}$
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