If θ1,θ2 are solutions of acos⁡2θ+bsin⁡2θ=c, tan⁡θ1≠tan⁡θ2 and a+c≠0, then tan⁡θ1+θ2=

acos⁡2θ+bsin⁡2θ=ca1−tan2⁡θ1+tan2⁡θ+b2tan⁡θ1+tan2⁡θ=c(c+a)tan2⁡θ−2btan⁡θ+c−a=0 roots tan⁡θ1,tan⁡θ2tan⁡θ1+tan⁡θ2=2bc+atan⁡θ,tan⁡θ2=c−ac+atan⁡θ1+θ2=2bc+a1−c−ac+a=ba

If θ1,θ2 are solutions of acos⁡2θ+bsin⁡2θ=c, tan⁡θ1≠tan⁡θ2 and a+c≠0, then tan⁡θ1+θ2=

acos⁡2θ+bsin⁡2θ=ca1−tan2⁡θ1+tan2⁡θ+b2tan⁡θ1+tan2⁡θ=c(c+a)tan2⁡θ−2btan⁡θ+c−a=0 roots tan⁡θ1,tan⁡θ2tan⁡θ1+tan⁡θ2=2bc+atan⁡θ,tan⁡θ2=c−ac+atan⁡θ1+θ2=2bc+a1−c−ac+a=ba

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $θ_{1}, θ_{2}$ are solutions of $acos 2 θ + bsin 2 θ = c, \tan θ_{1} \neq \tan θ_{2}$ and a+c $\neq$ 0, then $\tan (θ_{1} + θ_{2}) =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

-c/a

b/a

c/a

a/b

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} acos 2 θ + bsin 2 θ = c \\ a \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} + b \frac{2 \tan θ}{1 + \tan^{2} θ} = c \\ (c + a) \tan^{2} θ - 2 btan θ + c - a = 0 \\ roots \tan θ_{1}, \tan θ_{2} \end{array}$
$\begin{array}{l} \tan θ_{1} + \tan θ_{2} = \frac{2 b}{c + a} \\ \tan θ, \tan θ_{2} = \frac{c - a}{c + a} \\ \tan (θ_{1} + θ_{2}) = \frac{\frac{2 b}{c + a}}{1 - (\frac{c - a}{c + a})} = \frac{b}{a} \end{array}$