Question

If the roots of the equation (a - b)X + (b-c)x+ (c-a)=0 are equal,show thatc,a and b are in AP.​

Answer 1

Answer:

sinθ+cosθ=a

secθ+cscθ=b

\sf\underline \red{ To\:Find}

ToFind

We have to find the value of b(a²-1)

\sf\underline \pink{ Solution }

Solution

By putting the given values

\begin{gathered}:\implies\sf\ \ b(a^2-1)\\ \\ \\ :\implies\sf\ \ sec\theta+csc\theta\big\{(sin\theta+cos\theta)^2-1)\big\}\\ \\ \\ \bullet\sf\ sec\theta=\dfrac{1}{cos\theta}\ \ ;\ csc\theta=\dfrac{1}{sin\theta}\\ \\ \\ :\implies\sf\ \dfrac{1}{cos\theta}+\dfrac{1}{sin\theta}\big\{sin^2\theta+cos^2\theta+2sin\theta cos\theta-1\big\}\\ \\ \\ \bullet\sf\ \ sin^2\theta+cos^2\theta=1\\ \\ \\ :\implies\sf\dfrac{sin\theta+cos\theta}{sin\theta\ cos\theta}\big\{\cancel{1}+2sin\theta\ cos\theta \cancel{-1}\big\}\\ \\ \\ :\implies\sf\ \dfrac{sin\theta+cos\theta}{\cancel{sin\theta cos\theta}}\times 2\cancel{sin\theta cos\theta}\\ \\ \\ :\implies\sf\ \ 2(sin\theta+cos\theta)\end{gathered}

:⟹ b(a

2

−1)

:⟹ secθ+cscθ{(sinθ+cosθ)

2

−1)}

∙ secθ=

cosθ

1

; cscθ=

sinθ

1

:⟹

cosθ

1

+

sinθ

1

{sin

2

θ+cos

2

θ+2sinθcosθ−1}

∙ sin

2

θ+cos

2

θ=1

:⟹

sinθ cosθ

sinθ+cosθ

{

1

+2sinθ cosθ

−1

}

:⟹

sinθcosθ

sinθ+cosθ

×2

sinθcosθ

:⟹ 2(sinθ+cosθ)

\underline{\bigstar{\blue{\sf\ \ b(a^2-1)= 2(sin\theta+cos\theta)}}}

★ b(a

2

−1)=2(sinθ+cosθ