CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
In any $\triangle A B C$, if $a^2, b^2, c^2$ are in A.P., prove that $\cot A, \cot B$ and $\cot C$ are also in A.P.
✓
Answer
$a^2, b^2, c^2$ are in A.P.$\Rightarrow-2\text{a}^2,-2\text{b}^2,-2\text{c}^2,$ are in A.P.
$\Rightarrow(\text{a}^2+\text{b}^2+\text{c}^2)-2\text{a},(\text{a}^2+\text{b}^2+\text{c}^2)-2\text{b}^2,\$\text{a}^2+\text{b}^2+\text{c}^2)-2\text{c}^2$ are in A.P.
$\Rightarrow(\text{b}^2+\text{c}^2-\text{a}^2),(\text{c}^2+\text{a}^2-\text{b}^2),(\text{b}^2+\text{a}^2-\text{c}^2)$ are in A.P.
$\Rightarrow\frac{(\text{b}^2+\text{c}^2-\text{a}^2)}{2\text{abc}},\frac{(\text{c}^2+\text{a}^2-\text{b}^2)}{2\text{abc}},\frac{(\text{b}^2+\text{a}^2-\text{c}^2)}{2\text{abc}}$ are in A.P.
$\Rightarrow\frac{1}{\text{a}}\frac{(\text{b}^2+\text{c}^2-\text{a}^2)}{2\text{bc}},\frac{1}{\text{b}}\frac{(\text{c}^2+\text{a}^2-\text{b}^2)}{2\text{ac}},\frac{1}{\text{c}}\frac{(\text{b}^2+\text{a}^2-\text{c}^2)}{2\text{ab}}$ are in A.P.
$\Rightarrow\frac{1}{\text{a}}\cos\text{A},\frac{1}{\text{b}}\cos\text{B},\frac{1}{\text{c}}\cos\text{C}$ are in A.P.
$\Rightarrow\frac{\text{k}}{\text{a}}\cos\text{A},\frac{\text{k}}{\text{b}}\cos\text{B},\frac{\text{k}}{\text{c}}\cos\text{C}$ are in A.P.
$\Rightarrow\frac{\cos\text{A}}{\sin\text{A}},\frac{\cos\text{B}}{\sin\text{B}},\frac{\cos\text{C}}{\sin\text{C}}$ are in A.P.
$\Rightarrow\cot\text{A},\cot\text{B},\cot\text{C}$ are in A.P.
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