MCQ
If $\tan\text{A}+\cot\text{A}=4,$ then $\tan^4\text{A}+\cot^4\text{A}$ is equal to:
- A110
- B191
- C80
- D194
Solution:
We have:
$\tan\text{A}+\cot\text{A}=4$
squaring both the sides:
$(\tan\text{A}+\cot\text{A})^2=4^2$
$\Rightarrow\tan^2\text{A}+\cot^2\text{A}+2(\tan\text{A})(\cot\text{A})=16$
$\Rightarrow\tan^2\text{A}+\cot^2\text{A}+2=16$
$\Rightarrow\tan^2\text{A}+\cot\text{A}=14$
squaring both the sides again:
$(\tan^2\text{A}+\cot^2\text{A})^2=14^2$
$\tan^4\text{A}+\cot^4\text{A}+2(\tan^2\text{A})(\cot^2\text{A})=196$
$\Rightarrow\tan^4\text{A}+\cot^4\text{A}+2=196$
$\Rightarrow\tan^4\text{A}+\cot^4\text{A}=194$
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The equations of the lines which pass through the point (3, -2) and are inclined at 60° to the line $\sqrt{3} \text{x} + \text{y} = 1$ is:
$\sqrt{3}\text{x}+\text{y}-\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0$
$\sqrt{3}\text{x}+\text{y}+\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}+\sqrt{3}=0$
$\text{x}+\sqrt{3}\text{y}-\sqrt{3}=0,\text{x}-\sqrt{3}\text{y}-\sqrt{3}=0$