Question

If alpha and beta be two distinct real numbera satisfying the equation a cos x+b sin x =c then prove that tan (alpha +beta )= 2ab/a^2 - b^2

Answer 1

★ TRIGONOMETRIC REDUCTIONS ★

GIVEN EQUATION : acosx + bsinx = c
Provided roots of the equation - α and β

HENCE , acosα + bsinα = c , acosβ + bsinβ = c

Subtracting the subsequent second equation from the first , we obtain ;

a(cosα - cosβ) + b(sinα - sinβ) = 0
or ,
b ( sinα - sinβ ) - a ( cosβ - cosα ) = 0
or ,
2b cos α + β/2 sin α - β/2 = 2asin α + β/2 sin α- β /2
or ,
Tan α+ β /2 =b/a [ because α , β are different angles , hence can be substituted as sinα-β/2≠0

Hence , Sin (α+β)= 2Tan α+β/2 / 1 + Tan² α+β /2
= 2 { b/a} / 1 + b²/a² = 2ab / a² + b²

Hence , by further reductions ,

Tan (α+β)=2Tan(α+β/2)/1-Tan²( α+β /2 ) = 2ab / a² - b²


HENCE PROVED

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Answer 2

Given,

$a\cos\theta+b\sin\theta=c\\\;\\\text{Taking}\;\cos\theta\;\text{common}}\\\;\\a+b\frac{\sin\theta}{\cos\theta}=\frac{c}{\cos\theta}\\\;\\a+b\tan\theta=c\sec\theta\\\;\\\text{Making square of both sides,}\\\;\\(a+b\tan\theta)^2=(c\sec\theta)^2\\\;\\a^2+b^2\tan^2\theta+2ab\tan\theta=c^2\sec^2\theta\\\;\\a^2+b^2\tan^2\theta+2ab\tan\theta=c^2(1+\tan^2\theta)\;\;\;\;\;\;\;\;(\sec^2\theta=1+\tan^2theta)\\\;\\a^2+b^2\tan^2\theta+2ab\tan\theta=c^2+c^2\tan^2\theta\\\;\\(b^2-c^2)\tan^2\theta+2ab\tan\theta+a^2-c^2=0$

This  is in a  form of  a  quadratic equation, whoes  roots  are α and β. If  we compare  this equation with standard from, We  find  that  x is  compared to tan∅. hence, We can say that  tanα and  tanβ are solutions  of  this quadratic  equation.

Now,

$\text{Sum of roots}=-\frac{2ab}{b^2-c^2}\\\;\\\tan\alpha+\tan\beta=-\frac{2ab}{b^2-c^2}\;\;\;\;\;\;\..................i)\\\;\\\;\\\text{Product of roots}=\frac{a^2-c^2}{b^2-c^2}\\\;\\\tan\alpha.\tan\beta=\frac{a^2-c^2}{b^2-c^2}\;\;\;\;\;..............ii)$

$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha.\tan\beta}\\\;\\\tan(\alpha+\beta)=\frac{\frac{-2ab}{b^2-c^2}}{1-\frac{a^2-c^2}{b^2-c^2}}\\\;\\\tan(\alpha+\beta)=\frac{-2ab}{b^2-c^2-a^2+c^2}\\\;\\\tan(\alpha+\beta)=\frac{2ab}{b^2-a^2}$