If acos 2 -b sin 2 -c has - and - as its roots- prove that -tan-2 b-a-c and -tan-2-a-c-b2-c2-a2-

It is given that acos 2θ+bsin 2θ=c ⇒ a(1 tan^2 θ/1+tan^2 θ )+b(2 tan θ/1+tan^2 θ )=c [∵ cos 2θ=1 tan^2 θ/1+tan^2 θ and sin2θ=2tan θ/1+tan^2 θ ] ⇒ a(1 tan^2 ...

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Byju's Answer

Standard XI

Mathematics

Range

If acos 2 θ+b...

Question

If $a c o s 2 θ + b s i n 2 θ = c$ has $α$ and $β$ as its roots, prove that $α + t a n β = \frac{2 b}{a + c}$ and $α - t a n β = \frac{2}{a + c} \sqrt{b^{2} - c^{2} + a^{2} .}$

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Solution

It is given that $a c o s 2 θ + b s i n 2 θ = c$

$\Rightarrow a (\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}) + b (\frac{2 t a n θ}{1 + t a n^{2} θ}) = c [∵ c o s 2 θ = \frac{1 - t a n^{2} θ}{1 + t a n^{2} θ} a n d s i n 2 θ = \frac{2 t a n θ}{1 + t a n^{2} θ}]$

$\Rightarrow a (1 - t a n^{2} θ) + 2 b t a n θ = c (1 + t a n^{2} θ)$

$\Rightarrow (a + c) t a n^{2} θ - 2 b t a n θ + (c - a) = 0$

which is a quadratic equation in $t a n θ$

$∴$ Sum of roots = $t a n α + t a n β = - (\frac{- 2 b}{a + c}) = \frac{2 b}{a + c}$

and product of roots = $t a n α . t a n β = \frac{(c - a)}{a + c}$

Now, We have LHS = $t a n α - t a n β = \sqrt{(t a n α + t a n β)^{2} - 4 t a n α t a n β}$

= $\sqrt{{(\frac{2 b}{a + c})}^{2} - 4 (\frac{c - a}{a + c})} = \frac{2}{(a + c)} \sqrt{b^{2} - (c - a) (c + a)}$

= $\frac{2}{(a + c)} \sqrt{b^{2} - (c^{2} - a^{2})} = \frac{2}{(a + c)} \sqrt{b^{2} - c^{2} + a^{2}} = R H S$

Hence proved.

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If $a c o s 2 θ + b s i n 2 θ = c$ has $α a n d β$ as its roots, that prove that

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If acos 2θ+bsin 2θ=c has α and β as its roots, prove that α+tanβ=2ba+c and α−tan β=2a+c√b2−c2+a2.

If $a c o s 2 θ + b s i n 2 θ = c$ has $α$ and $β$ as its roots, prove that $α + t a n β = \frac{2 b}{a + c}$ and $α - t a n β = \frac{2}{a + c} \sqrt{b^{2} - c^{2} + a^{2} .}$