UPSC CDS 1 2022 Math Paper

Shortcut Trick Given equation is $\tan^{2}\theta+3\sec\theta-9=0$ How to think about trick: Since R.H.S is a rational number So, L.H.S should be rational for this $3\sec\theta$ must be rational because $\tan^{2}\theta$ and 9 are rational number Now, we have to think that on which $\theta,\ \sec\theta$ is rational so we get $\theta=60^{\circ}$ We can't take $0^{\circ}$ or $90^{\circ}$ according to the question Process: Put $\theta=60^{\circ}$ in the given equation We get, LHS $=\tan^{2}60^{\circ}+3\sec60^{\circ}-9$ $\Rightarrow 3+3\times2-9=0=\;\text{RHS}\;$ Now, We have to find the value of $12\cot^{2}\theta+3\csc\theta$, at $\theta=60^{\circ}$ $\Rightarrow 12\cot^{2}60^{\circ}+\csc60^{\circ}$ $\Rightarrow 12\left(\frac{1}{3}\right)+3\times\frac{2}{\sqrt{3}}$ $\Rightarrow 4+2\sqrt{3}=(\sqrt{3})^{2}+(1)^{2}+2\cdot(\sqrt{3})\cdot(1)$ Using formula $a^{2}+b^{2}+2ab=(a+b)^{2}$ $\Rightarrow(\sqrt{3}+1)^{2}$ $\therefore$ The required value is $(\sqrt{3}+1)^{2}$