If a cos 2 theta- - b sin 2 theta- - c has alpha- and beta- as its roots- then prove that tan alpha- - tan beta- - 2b-a-c-

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Standard X

Mathematics

Range of Trigonometric Ratios from 0 to 90 Degrees

If a cos 2 ...

Question

If a cos 2θ + b sin 2θ = c has α and β as its roots, then prove that tanα + tan β = 2b/a+c
.

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Solution

$a \cos 2 θ + b \sin 2 θ = c a (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) + b (\frac{2 \tan θ}{1 + \tan^{2} θ}) = c (Since \sin 2 x = \frac{2 \tan x}{1 + \tan^{2} x} and \cos 2 x = \frac{1 - \tan^{2} x}{1 + \tan^{2} x}) a - a \tan^{2} θ + 2 b \tan θ = c + c \tan^{2} θ (c + a) \tan^{2} θ - 2 b \tan θ + (c - a) = 0 \tan θ = \frac{2 b \pm \sqrt{4 b^{2} - 4 (c + a) (c - a)}}{2 (c + a)} = \frac{2 b \pm \sqrt{4 b^{2} - 4 c^{2} + 4 a^{2}}}{2 (c + a)} = \frac{2 b \pm 2 \sqrt{b^{2} - c^{2} + a^{2}}}{2 (c + a)} = \frac{b \pm \sqrt{b^{2} - c^{2} + a^{2}}}{(c + a)} So, α = \tan^{- 1} (\frac{b + \sqrt{b^{2} - c^{2} + a^{2}}}{(c + a)}) and β = \tan^{- 1} (\frac{b - \sqrt{b^{2} - c^{2} + a^{2}}}{(c + a)}) \tan α + \tan β = \tan (\tan^{- 1} (\frac{b + \sqrt{b^{2} - c^{2} + a^{2}}}{(c + a)})) + \tan (\tan^{- 1} (\frac{b - \sqrt{b^{2} - c^{2} + a^{2}}}{(c + a)})) = (\frac{b + \sqrt{b^{2} - c^{2} + a^{2}}}{(c + a)}) + (\frac{b - \sqrt{b^{2} - c^{2} + a^{2}}}{(c + a)}) = \frac{b + \sqrt{b^{2} - c^{2} + a^{2}} + b - \sqrt{b^{2} - c^{2} + a^{2}}}{(c + a)} = \frac{2 b}{c + a} = RHS Hence Proved .$

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Q. If

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Q. If

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If a cos 2θ + b sin 2θ = c has α and β as its roots, then prove that tanα + tan β = 2b/a+c .

If a cos 2θ + b sin 2θ = c has α and β as its roots, then prove that tanα + tan β = 2b/a+c
.