Question

$(m+2)\sin \theta +2(m-1)\cos \theta =2m+1$. If

• A. $\tan \theta =\frac{3}{4}$
• B. $\tan \theta =\frac{4}{3}$
• C. $\tan \theta =2m/(m^2-1)$
• D. $\tan \theta =2m/(m^2+1)$

Answer 1

Answer: (A), (C)

The correct options are
A $\tan \theta =\frac{3}{4}$
C $\tan \theta =2m/(m^2-1)$

The given relation can be written as

$(m+2)\tan \theta +(2m-1)=(2m+1)sec\theta \Rightarrow (m+2{)}^2{\tan }^2\theta +2(m+2\left)\right(2m-1)\tan \theta +(2m-1{)}^2=(2m+1{)}^2(1+{\tan }^2\theta )\Rightarrow \left[\right(m+2{)}^2-(2m+1{)}^2{\tan }^2\theta ]+2(m+2\left)\right(2m-1)\tan \theta +(2m-1{)}^2-(2m+1{)}^2=0\Rightarrow 3(1-m^2){\tan }^2\theta +(4m^2+6m-4)\tan \theta -8m=0\Rightarrow (3\tan \theta -4)\left[\right(1-m^2)\tan \theta +2m]=0$

Which is true as $\tan \theta =\frac{4}{3}$ or $\tan \theta =2m/(m^2-1)$

Therefore, Answer is $\tan \theta =\frac{4}{3}$ or $\tan \theta =2m/(m^2-1)$