Question

If 6 sin²θ + 2 cos²θ=7 sinθcosθ then find the value of tan θ. Class 10

Answer 1

$\textbf{Given:}$

$6\,sin^2\theta+2\,cos^2\theta=7\,sin\theta\,cos\theta$

$\textbf{To find:}$

$\text{The value of}\;\tan\theta$

$\textbf{Solution:}$

$\text{Consider,}$

$6\,sin^2\theta+2\,cos^2\theta=7\,sin\theta\,cos\theta$

$\text{Divide bothsides by \sin\theta\,\cos\theta }$

$\dfrac{6\,sin^2\theta}{sin\theta\,cos\theta}+\dfrac{2\,cos^2\theta}{sin\theta\,cos\theta}=7$

$\dfrac{6\,sin\theta}{\cos\theta}+\dfrac{2\,cos\theta}{sin\theta}=7$

$6\,tan\theta+2\,cot\theta=7$

$6\,tan\theta+2(\dfrac{1}{tan\theta})=7$

$6\,tan^2\theta+2=7\,tan\theta$

$6\,tan^2\theta+7\,tan\theta+2=0$

$6\,tan^2\theta+4\,tan\theta+3\,tan\theta+2=0$

$2\,tan\theta(3\,tan\theta+2)+1(3\,tan\theta+2)=0$

$(2\,tan\theta+1)(3\,tan\theta+2)=0$

$\implies\,tan\theta=\dfrac{-1}{2},\dfrac{-2}{3}$

$\textbf{Answer:}$

$\text{The values of tan\theta are \dfrac{-1}{2},\,\dfrac{-2}{3} }$.