If acos 2- -bsin 2- -c has - and - as its solutions -then show that tan- -tan- -2b-c-a- tan-tan- -c-a-c-a-

acos 2θ #160; +bsin 2θ #160; =c⇒ acos 2θ #160; =c bsin 2θ #160; #160;converting it into quadratic equation of sin2θ ⇒ a ^ 2 ( 1 sin ^ ...

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Byju's Answer

Standard XII

Mathematics

Modulus of a Complex Number

If acos 2θ ...

Question

If $a cos 2 θ + b sin 2 θ = c$ has $α$ and $β$ as its solutions ,then show that $tan α + tan β = \frac{2 b}{c + a}, tan α tan β = \frac{c - a}{c + a}$ .

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Solution

$a cos 2 θ + b sin 2 θ = c \Rightarrow a cos 2 θ = c - b sin 2 θ$

converting it into quadratic equation of $s i n 2 θ$

$\Rightarrow a^{2} (1 - {sin}^{2} 2 θ) = c^{2} + b^{2} {sin}^{2} 2 θ - 2 b c sin 2 θ \Rightarrow (a^{2} + b^{2}) {sin}^{2} 2 θ - 2 b c sin 2 θ + (c^{2} - a^{2}) = 0$

using the properties of roots of solution of quadratic equation

$\Rightarrow sin 2 α + sin 2 β = \frac{2 b c}{a^{2} + b^{2}}$ and $sin 2 α sin 2 β = \frac{2 b c}{c^{2} - a^{2}}$

converting it into quadratic equation of $c o s 2 θ$

$a cos 2 θ + b sin 2 θ = c \Rightarrow b sin 2 θ = c - a cos 2 θ \Rightarrow b^{2} (1 - {cos}^{2} 2 θ) = c^{2} + a^{2} {cos}^{2} 2 θ - 2 a c cos 2 θ = 0 \Rightarrow (a^{2} + b^{2}) {cos}^{2} 2 θ - 2 a c cos 2 θ + (c^{2} + b^{2}) = 0$

using the properties of roots of solution of quadratic equation

$∴ cos 2 α + cos 2 β = \frac{2 a c}{a^{2} + b^{2}}$ and $cos 2 α cos 2 β = \frac{2 a c}{c^{2} + b^{2}}$

$∴ tan α + tan β = \frac{2 b}{c + a}$ and $tan α tan β = \frac{c - a}{c + a}$

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Similar questions

If acos2 $θ$ + bsin2 $θ$ = C has $α$ and $β$ as its solution, then values of tan $α$ + tan $β$ ,tan $α$ .tan $β$ are respectivly equal to

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

(i) $t a n α + t a n β = \frac{2 b}{a + c}$

(ii) $t a n α t a n β = \frac{c - a}{c + a}$

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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 $α - t a n β = \frac{2}{a + c} \sqrt{b^{2} - c^{2} + a^{2} .}$

Q. 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

Q. If

a \cos 2 x + b \sin 2 x = c

has α and β as its roots, then prove that

(i)

\tan α + \tan β = \frac{2 b}{a + c}

[NCERT EXEMPLAR]

(ii)

\tan α \tan β = \frac{c - a}{c + a}

(iii)

\tan (α + β) = \frac{b}{a}

[NCERT EXEMPLAR]

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If acos2θ+bsin2θ=c has α and β as its solutions ,then show that tanα+tanβ=2bc+a,tanαtanβ=c−ac+a.

If $a cos 2 θ + b sin 2 θ = c$ has $α$ and $β$ as its solutions ,then show that $tan α + tan β = \frac{2 b}{c + a}, tan α tan β = \frac{c - a}{c + a}$ .