If - - are the solution of a cos 2 - b sin 2 - c- then tan-tan-

The correct option is B a cos 2θ +b sin 2θ =c ⇒ a ( 1 tan^2 θ/1 tan^2 θ )+ ( 2 tan θ/1+tan^2 θ )=c ⇒ a(1 tan^2 θ)+2b tan θ =c (1+tan^2 θ) ⇒ (a+c)tan^2 θ 2b t ...

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Byju's Answer

Standard XII

Mathematics

Equations of the Form acosθ+bsinθ = c

If α, β are t...

Question

If $α, β$ are the solution of a $c o s 2 θ + b$ $s i n 2 θ = c$ , then $t a n α t a n β$

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Solution

The correct option is C
$a c o s 2 θ + b s i n 2 θ = c \Rightarrow a (\frac{1 - t a n^{2} θ}{1 - t a n^{2} θ}) + (\frac{2 t a n θ}{1 + t a n^{2} θ}) = c \Rightarrow a (1 - t a n^{2} θ) + 2 b t a n θ = c (1 + t a n^{2} θ)$
$\Rightarrow (a + c) t a n^{2} θ - 2 b t a n θ + c - a = 0 ∴ t a n α . t a n β = \frac{c - a}{c + a}$

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If α,β are the solution of a cos2θ+b sin2θ=c, then tanαtanβ

The correct option is C a cos2θ+bsin2θ=c⇒a(1−tan2θ1−tan2θ)+(2tanθ1+tan2θ)=c⇒a(1−tan2θ)+2btanθ=c(1+tan2θ) ⇒(a+c)tan2θ−2btanθ+c−a=0∴tanα.tanβ=c−ac+a

If $α, β$ are the solution of a $c o s 2 θ + b$ $s i n 2 θ = c$ , then $t a n α t a n β$