Question

Question 4
If $1+si{n}^2\theta =3sin\theta .cos\theta$, then prove that $tan\theta =1$ or $\frac{1}{2}$.

Answer 1

Given, $1+si{n}^2\theta =3sin\theta .cos\theta$

On dividing by $si{n}^2\theta$ on both sides, we get;

$\frac{1}{si{n}^2\theta }+1=3.cot\theta$ $[\because cot\theta =\frac{cos\theta }{sin\theta }]$

$\Rightarrow cose{c}^2\theta +1=3.cot\theta$ $[cosec\theta =\frac{1}{sin\theta }]$

$\Rightarrow 1+co{t}^2\theta +1=3.cot\theta$ $\because cose{c}^2\theta -co{t}^2\theta =1$

$\Rightarrow co{t}^2\theta -3cot\theta +2=0$

$\Rightarrow co{t}^2\theta -2cot\theta -cot\theta +2=0$ [by splitting the middle term]

$\Rightarrow cot\theta (cot\theta -2)-1(cot\theta -2)=0$

$\Rightarrow (cot\theta -2)(cot\theta -1)=0\Rightarrow cot\theta =1or2$

$\Rightarrow tan\theta =1or\frac{1}{2}$ $[\because tan\theta =\frac{1}{cot\theta }]$