Question

If $1+{\sin }^2\mathrm{\theta }=3\mathrm{sin\theta cos\theta }$ then prove that $\mathrm{tan\theta }=1$or $\frac{1}{2}$

Answer 1

Step 1: Divide both sides of given equation by ${\sin }^2\mathrm{\theta }$

Given:$1+{\sin }^2\mathrm{\theta }=3\mathrm{sin\theta cos\theta }$

$\frac{1}{{\sin }^2\mathrm{\theta }}+\frac{{\sin }^2\mathrm{\theta }}{{\sin }^2\mathrm{\theta }}=\frac{3\mathrm{sin\theta cos\theta }}{{\sin }^2\mathrm{\theta }}$

$\Rightarrow$ $1+\frac{1}{{\sin }^2\mathrm{\theta }}=\frac{3\mathrm{cos\theta }}{\mathrm{sin\theta }}$

$\Rightarrow$ $1+{\mathrm{cosec}}^2\mathrm{\theta }=3\mathrm{cot\theta }$ $...\left(\mathrm{i}\right)$

Step 2: Use trigonometric identities to reduce equation $\left(\mathrm{i}\right)$ to an equation in $\mathrm{cot\theta }$

we know that ${\mathrm{cosec}}^2\mathrm{\theta }={\cot }^2\mathrm{\theta }+1$

$\Rightarrow$ $1+{\cot }^2\mathrm{\theta }+1=3\mathrm{cot\theta }$

$\Rightarrow$ ${\cot }^2\mathrm{\theta }-3\mathrm{cot\theta }+2=0$ $...\left(\mathrm{ii}\right)$

Step 3: Solve equation $\left(\mathrm{iii}\right)$ to obtain value of $\mathrm{cot\theta }$ and consequently $\mathrm{tan\theta }$

${\cot }^2\mathrm{\theta }-3\mathrm{cot\theta }+2=0$

$\Rightarrow$ ${\cot }^2\mathrm{\theta }-\mathrm{cot\theta }-2\mathrm{cot\theta }+2=0$

$\Rightarrow$ $\mathrm{cot\theta }\left(\mathrm{cot\theta }-1\right)-2\left(\mathrm{cot\theta }-1\right)=0$

$\Rightarrow$ $\left(\mathrm{cot\theta }-1\right)\left(\mathrm{cot\theta }-2\right)=0$

$\Rightarrow$ $\mathrm{cot\theta }=1$ or $\mathrm{cot\theta }=2$

we know $\mathrm{tan\theta }=\frac{1}{\mathrm{cot\theta }}$

$\Rightarrow$ $\mathrm{tan\theta }=1$ or $\mathrm{tan\theta }=\frac{1}{2}$

Hence, proved that $\mathrm{tan\theta }=1$ or $\mathrm{tan\theta }=\frac{1}{2}$