If a cos 2-b sin 2-c has - and - as its solution- then the value of tan - - tan - is

A cos 2θ+b sin 2θ=c ⇒ a (1 tan^2 θ/1+tan^2 θ)+b 2 tan θ/1+tan^2 θ=c ⇒ a atan^2 θ+2b tan θ=c+c tan^2 θ ⇒ (a+c)tan^2 θ+2b tan θ+(a c)=0 ∴ tan α + tan β= 2b/ (c+a ...

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Byju's Answer

Standard XIII

Mathematics

Trigonometric Ratios of Compound Angles

If a cos 2θ+b...

Question

If $a c o s 2 θ + b s i n 2 θ = c h a s α a n d β$
as its solution, then the value of $t a n α + t a n β$ is

\frac{c + a}{2 b}

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\frac{2 b}{c + a}

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\frac{c - a}{2 b}

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\frac{b}{c + a}

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Solution

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

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If a cos 2θ+b sin 2θ=c has α and β as its solution, then the value of tan α+tan β is

The correct option is B 2bc+aa cos 2θ+b sin 2θ=c ⇒a(1−tan2 θ1+tan2 θ)+b2−tan θ1+tan2 θ=c ⇒a−atan2 θ+2b tan θ=c+c tan2 θ ⇒−(a+c)tan2 θ+2b tan θ+(a−c)=0 ∴tan α+tan β=−2b−(c+a)=2bc+a.

If $a c o s 2 θ + b s i n 2 θ = c h a s α a n d β$
as its solution, then the value of $t a n α + t a n β$ is