Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Trignometry questions?
Postings ome questions from another site
1. Solve : 2 tan x = 4x - 2 sin 2x cos 2x tan² x
2. Show that the solutions of the equation tan ax = tan bx, a² + b² <> 0 are in A.P. Find the common difference of the series.
3. If α , β be the roots of the equation a cos 2θ + b sin 2θ =c then show that
tan α + tan β = 2b / (c + a) and tan α tan β = (c - a) / (c + a)
2 Answers
- MadhukarLv 710 years agoFavorite Answer
2.
tanax = tanbx
=> ax = kπ + bx, k ∈ Z - {0}
=> x = kπ/(a - b), k ∈ Z - {0}
=> x = { ..... -2π(a - b), - π/(a - b), π/(a - b), 2π/(a - b), .....}
=> the solutions for a^2 + b^2 ≠ 0 are in A.P.
and common difference = π/(a - b)
3.
a cos 2θ + b sin 2θ = c
=> a (1 - tan^2 θ) / (1 + tan^2 θ) + 2b tan θ / (1 + tan^2 θ) = c
=> a (1 - tan^2 θ) + 2b tan θ = c (1 + tan^2 θ)
=> (c + a) tan^2 θ - 2b tan θ + (c - a) = 0
If α , β be the roots of the equation which is a quadratic in tan θ,
sum of the roots
= tan α + tan β = 2b/(c + a)
and product of the roots,
tan α * tan β = (c - a) / (c + a).
- RameshwarLv 710 years ago
3. acos2x + bsin2x = c [ take theta = x ]
a + b tan2 x = c/cos2x
a + b [2tanx /1- tan^2x ] = c [1+tan^2x/1- tan^2x]
a [ 1- tan^2x] + b [2tanx] = c[ 1+ tan^2x]
a - atan^2x + 2b tanx = c + c tan^2x
tan^2x [a+c ] -- 2b tanx + {c-a} =0
now as tan alpha & tan beta are roots of this equation
so some of roots , i.e. tan alpha + tan beta = 2b/(c+a ) PROVED
and product of roots i. e. tan alpha * tan beta = c-a / c+a PROVED