Question

If a cos 2θ + b sin 2θ = c has α and β as its roots, then prove that tan α + tan β = 2b/(a + c).
Note
Do not copy as it is easy

Answer 1

Answer:

Step-by-step explanation:

Answer: (2) 2b/(c + a)

Solution:

Given,

a cos 2θ + b sin 2θ = c….(i)

a[(1 – tan2θ)/(1 + tan2θ)] + b[2 tan θ/(1 + tan2θ)] = c

a(1 – tan2θ) + 2b tan θ = c(1 + tan2θ)

c + c tan2θ – a + a tan2θ – 2b tan θ = 0

(c + a)tan2θ – 2b tan θ + (c – a) = 0….(ii)

Given that α and β are the roots of equation (i).

Thus, tan α and tan β are the roots of equation (ii) as it is a quadratic equation in tan θ.

Sum of the roots of equation (ii),

tan α + tan β = -(-2b)/(c + a) = 2b/(c + a)

the sum of the lower limits  of the median and modal class is

MARK ME AS BRAINLIEST!!!

HAPPY TO HELP(●‿●)

FOLLOW

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: \alpha , \beta \: are \: roots \: of \: acos2 \theta \: + \: bsin2\theta = c$

Now, we have the equation,

$\rm :\longmapsto\: \: acos2 \theta \: + \: bsin2\theta = c$

We know that,

$\underbrace{\boxed{ \tt{ sin2x\: = \: \frac{2tanx}{1 + {tan}^{2}x }}}}$

and

$\underbrace{\boxed{ \tt{ cos2x\: = \: \frac{1 - {tan}^{2}x }{1 + {tan}^{2}x }}}}$

So,

On substituting the values, in given equation, we get

$\rm :\longmapsto\:a\bigg(\dfrac{1 - {tan}^{2} \theta }{1 + {tan}^{2} \theta } \bigg) + b\bigg(\dfrac{2tan\theta }{1 + {tan}^{2} \theta } \bigg) = c$

On taking LCM, we get

$\rm :\longmapsto\:a(1 - {tan}^{2}\theta ) + 2 b \: tan\theta = c(1 + {tan}^{2}\theta )$

$\rm :\longmapsto\:a - a{tan}^{2}\theta + 2 b \: tan\theta = c + c{tan}^{2}\theta$

can be rewritten as

$\rm :\longmapsto\:a {tan}^{2}\theta + c {tan}^{2}\theta - 2 btan\theta - a + c = 0$

$\rm :\longmapsto\:(a + c) {tan}^{2}\theta - 2 btan\theta + (c - a) = 0$

Now, its a quadratic in tan,

So,

$\rm :\implies\:Roots \: of \: the \: equation \: are \: tan \alpha \: and \: tan \beta$

We know that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

Thus,

$\rm :\longmapsto\: tan \alpha + tan \beta \: = \: - \: \dfrac{ (- 2b)}{a + c}$

$\bf\implies \:\: tan \alpha + tan \beta \: = \: \: \dfrac{2b}{ \: \: a + c \: \: }$

Hence, Proved

Additional Information :-

$\boxed{ \sf{ \:sin2x = 2sinxcosx}}$

$\boxed{ \sf{ \:cos2x = 1 - {2sin}^{2}x}}$

$\boxed{ \sf{ \:cos2x = {2cos}^{2}x - 1}}$

$\boxed{ \sf{ \:cos2x = {cos}^{2}x - {sin}^{2}x}}$

$\boxed{ \sf{ \:tan2x = \frac{2tanx}{1 - {tan}^{2} x}}}$

$\boxed{ \sf{ \:sin3x = 3sinx - {4sin}^{3}x}}$

$\boxed{ \sf{ \:cos3x = {4cos}^{3}x - 3cosx}}$