Question

$(m+2)sin\theta +(2m-1)cos\theta =2m+1$, if

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

Answer 1

Answer: (B), (C)

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

$(m+2)sin\theta +(2m-1)cos\theta =(2m+1)$
Dividing both sides by $cos\theta$, we get
$(m+2)tan\theta +(2m-1)=(2m+1)sec\theta$
Squaring both sides, we get
$(m+2{)}^{2}ta{n}^2\theta +2(m+2\left)\right(2m-1)tan\theta +(2m-1{)}^2=(2m+1{)}^2(1+ta{n}^2\theta )\Rightarrow [(m+2{)}^2-(2m-1{)}^2]ta{n}^2\theta +2(m+2\left)\right(2m-1)tan\theta +(2m-1{)}^2-(2m+1{)}^2=0\Rightarrow 3(1-m^2\left)ta{n}^2\theta \right(4m^2+6m-4)tan\theta -8m=0\Rightarrow (3tan\theta -4\left)\right[(1-m^2)tan\theta +2m]=0$
Therefore,
$3tan\theta =4,1-m^2)tan\theta +2m=0\Rightarrow tan\theta =\frac{4}{3},tan\theta =\frac{2m}{m^2-1}$
Hence, options 'B' and 'C' are correct.