Question

If $3{\cos }^2\theta +7{\sin }^2\theta =4$, show that $\cot \theta =\sqrt{3}$.

Answer 1

Given,

$3{\cos }^2\theta +7{\sin }^2\theta =4$

or, $3(1-{\sin }^2\theta )+7{\sin }^2\theta =4$

or, $4{\sin }^2\theta =1$

or, $cose{c}^2\theta =4$.......(1).

We have,

$\cot \theta =\sqrt{-1+cose{c}^2\theta }=\sqrt{-1+4}=\sqrt{3}$. [ Using (1)]