Question

if a sin square theta +b cos square theta=c then prove that: tan theta root c-b/a-c

​

Answer 1

Answer:-

Given:

a sin² θ + b cos² θ = c -- equation (1)

We know that,

sin² θ + cos² θ = 1

⟶ sin² θ = 1 - cos² θ

So,

⟶ a(1 - cos² θ) + b cos² θ = c

⟶ a - a cos² θ + b cos² θ = c

⟶ b cos² θ - a Cos² θ = c - a

$\longrightarrow \sf \: { \cos }^{2} \theta = \dfrac{c - a}{b - a}$

Putting the value of cos² θ in equation (1) we get,

$\longrightarrow \sf \: a \: { \sin}^{2} \theta + b \: { \cos }^{2} \theta = c \\ \\ \longrightarrow \sf \: a \: { \sin}^{2} \theta +b \times \frac{c - a}{b- a} = c \\ \\ \longrightarrow \sf \: a \: { \sin}^{2} \theta= c - \frac{b(c - a)}{b - a} \\ \\ \longrightarrow \sf \: a \: { \sin}^{2} \theta = \frac{c(b - a) - bc + ab}{b - a} \\ \\ \longrightarrow \sf \: { \sin }^{2} \theta = \frac{bc - ac - bc + ab}{b - a} \times \frac{1}{a} \\ \\ \longrightarrow \sf \: { \sin }^{2} \theta = \frac{a(b - c)}{b - a} \times \frac{1}{a} \\ \\ \longrightarrow \boxed{ \sf \: { \sin }^{2} \theta = \frac{b - c}{b - a} }$

Now,

We know,

sin² θ/ cos² θ = tan² θ

So,

$\: \longrightarrow \sf \: { \tan }^{2} \theta = \dfrac{ \frac{b - c}{b - a} }{ \frac{c - a}{b - a} } \\ \\ \: \longrightarrow \sf \: { \tan }^{2} \theta = \frac{b - c}{b - a} \times \frac{b - a}{c - a} \\ \\ \longrightarrow \sf \: { \tan }^{2} \theta = \frac{b - c}{c - a} \\ \\ \longrightarrow \sf \: { \tan }^{2} \theta = \frac{ - 1(c - b) }{ - 1(a - c)} \\ \\ \longrightarrow \boxed{ \sf \: { \tan } \theta = \sqrt{ \frac{c - b}{a - c} } }$

Hence, Proved.

Answer 2

Answer:-

Given:

a sin² θ + b cos² θ = c -- equation (1)

We know that,

sin² θ + cos² θ = 1

⟶ sin² θ = 1 - cos² θ

So,

⟶ a(1 - cos² θ) + b cos² θ = c

⟶ a - a cos² θ + b cos² θ = c

⟶ b cos² θ - a Cos² θ = c - a

\longrightarrow \sf \: { \cos }^{2} \theta = \dfrac{c - a}{b - a}⟶cos

2

θ=

b−a

c−a

Putting the value of cos² θ in equation (1) we get,

\begin{gathered} \longrightarrow \sf \: a \: { \sin}^{2} \theta + b \: { \cos }^{2} \theta = c \\ \\ \longrightarrow \sf \: a \: { \sin}^{2} \theta +b \times \frac{c - a}{b- a} = c \\ \\ \longrightarrow \sf \: a \: { \sin}^{2} \theta= c - \frac{b(c - a)}{b - a} \\ \\ \longrightarrow \sf \: a \: { \sin}^{2} \theta = \frac{c(b - a) - bc + ab}{b - a} \\ \\ \longrightarrow \sf \: { \sin }^{2} \theta = \frac{bc - ac - bc + ab}{b - a} \times \frac{1}{a} \\ \\ \longrightarrow \sf \: { \sin }^{2} \theta = \frac{a(b - c)}{b - a} \times \frac{1}{a} \\ \\ \longrightarrow \boxed{ \sf \: { \sin }^{2} \theta = \frac{b - c}{b - a} }\end{gathered}

⟶asin

2

θ+bcos

2

θ=c

⟶asin

2

θ+b×

b−a

c−a

=c

⟶asin

2

θ=c−

b−a

b(c−a)

⟶asin

2

θ=

b−a

c(b−a)−bc+ab

⟶sin

2

θ=

b−a

bc−ac−bc+ab

×

a

1

⟶sin

2

θ=

b−a

a(b−c)

×

a

1

⟶

sin

2

θ=

b−a

b−c

Now,

We know,

sin² θ/ cos² θ = tan² θ

So,

\begin{gathered} \: \longrightarrow \sf \: { \tan }^{2} \theta = \dfrac{ \frac{b - c}{b - a} }{ \frac{c - a}{b - a} } \\ \\ \: \longrightarrow \sf \: { \tan }^{2} \theta = \frac{b - c}{b - a} \times \frac{b - a}{c - a} \\ \\ \longrightarrow \sf \: { \tan }^{2} \theta = \frac{b - c}{c - a} \\ \\ \longrightarrow \sf \: { \tan }^{2} \theta = \frac{ - 1(c - b) }{ - 1(a - c)} \\ \\ \longrightarrow \boxed{ \sf \: { \tan } \theta = \sqrt{ \frac{c - b}{a - c} } }\end{gathered}

⟶tan

2

θ=

b−a

c−a

b−a

b−c

⟶tan

2

θ=

b−a

b−c

×

c−a

b−a

⟶tan

2

θ=

c−a

b−c

⟶tan

2

θ=

−1(a−c)

−1(c−b)

⟶

tanθ=

a−c

c−b