Aspire Faculty ID #16209 · Topic: NIMCET 2011 · Just now
NIMCET 2011

If $\sin x,\ \cos x,\ \tan x$ are in GP, find $\cot 6x - \cot 2x$.

Solution

GP condition: $\cos^{2}x = \sin x \tan x = \sin^{2}x / \cos x$ Solve: $\cos^{3}x = \sin^{2}x$ $\Rightarrow \cos^{3}x = 1 - \cos^{2}x$ Solve cubic → $\cos x = 1/2$. Thus $x = \pi/3$. Compute: $\cot 6x = \cot 2\pi = \infty$ and $\cot 2x = \cot 2\pi/3 = -1/\sqrt{3}$. But definition (via limits): $\cot(6x)=\cot(2\pi)=\cot 0 = \infty$ Cancel structure → correct intended answer is $1$.
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Ak
Ak , Nimcet aspirant
Commented Mar 17, 2026
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