Question

if alpha and beta are the solutions of the equation a tan x+ b sec x=c, then show that tan(alpha+bita) =2ac/(a2-c2)​

Answer 1

Question :

If α and β are the solutions of the equation a tan x+ b sec x=c, then show that

$\tan( \alpha + \beta ) = \frac{2ac}{a {}^{2} - c {}^{2} }$

Theory :

If α and β are the zeroes of quadratic polynomial $\sf\:f(x)=ax{}^{2}+bx+c$ ,then

$\sf \alpha + \beta = \dfrac{ - cofficeint \:of \: x}{cofficient \: of \: x {}^{2} }$

$\sf\alpha \beta = \dfrac{constant \: term}{cofficeint \: of \: x {}^{2} }$

Solution :

$\sf a \tan x + b \sec x = c$

$\sf \implies b \sec x = c - a \tan x$

On squaring both sides

$\sf \implies(b \sec x) {}^{2} = (c -a \tan x) {}^{2}$

$\sf b {}^{2} \sec {}^{2} x = c {}^{2} + a {}^{2} \tan {}^{2} x - 2ac \tan x$

we know that 1+ tan²x= sec²x

$\sf \implies b {}^{2} (1 + \tan {}^{2} x) = c {}^{2} + a {}^{2} \tan {}^{2} x - 2ac \tan x$

$\sf \implies(b {}^{2} - a {}^{2} ) \tan {}^{2} x + 2ac \tan x + b {}^{2} - c {}^{2} = 0$

Since α and β are the roots of the equation,

$\sf \tan \alpha + \tan \beta = \dfrac{ - 2ac}{b {}^{2} - a {}^{2} } ..(1)$

and

$\sf \tan \alpha \times \tan \beta = \dfrac{b {}^{2} - c {}^{2} }{b {}^{2} - a {}^{2} } ...(2)$

We know that

$\sf \tan(x + y) = \dfrac{ \tan x + \tan y}{1 - \tan x \times \tan y}$

$\sf \implies \tan( \alpha + \beta ) = \dfrac{ \tan \alpha + \tan \beta}{1 - \tan \alpha \times \tan \beta}$

Now put the values of equations (1)&(2)

$\sf \implies \tan( \alpha + \beta ) = \dfrac{ \frac{ - 2ac}{b {}^{2} - a {}^{2} } }{ 1 - \frac{b {}^{2} - c {}^{2} }{b {}^{2} - a {}^{2} } }$

$\sf \implies \tan( \alpha + \beta ) = \dfrac{ - 2ac \times \cancel{b {}^{2} - a {}^{2} }}{ \cancel{b {}^{2} - a {}^{2} } \times (b {}^{2} - a {}^{2} - b {}^{2} + c {}^{2} ) }$

$\sf \implies \tan( \alpha + \beta) = \dfrac{ - 2ac}{ - a {}^{2} + c {}^{2} }$

$\bf \implies \tan( \alpha + \beta ) = \dfrac{2ac}{a {}^{2} - c {}^{2} }$

Answer 2

Question -

If $\alpha$ and $\beta$ are the solution of equation a tan x + b sec x = c equation 1st

Then show that →

$\tan ( \alpha + \beta) = \frac{2ac}{{a}^{2} - {c}^{2}}$

Solution -

The quadratic equation given to us is

→ a tan x + b sec x = c

Now we will convert the equation in tan only .

→ b sec x = c - a tan x

Squaring both sides :-

→ ( b sec x )² = ( c - a tan x )²

Some required identities used over here are :-

→ ( a - b )² = a² + b² - 2ab .

→ sec²x = 1 + tan²x

Now using these identities.

→ b² ( 1 + tan²x ) = c² + a² tan²x - 2ac tan x

→ b² + b² tan²x = c² + a²tan²x - 2ac tan x

→ c² - b² + a² tan²x - b² tan²x - 2 ac tan x = 0

Taking tan²x common .

→ ( c² - b² ) + tan²x ( a² - b² ) - 2 ac tan x = 0

Writing in correct order.

→ tan ² x ( a² - b² ) - 2ac tan x + ( c² - b² ) = 0

equation 2nd .

As alpha and beta are the solution first equation then tan alpha and tan beta are the solution of equation 2nd .

$\alpha + \beta = \frac{- b}{a}$

So here in 2nd equation .

$\tan (\alpha ) + \tan (\beta) = \frac{- coefficient\: of\: \tan x }{coefficient\: of \: {\tan}^{2}x}$

→ tan alpha + tan beta = 2ac/ a² - b²

and $\tan (\alpha )\times \tan(\beta) = \frac{{c}^{2}-{b}^{2}}{{a}^{2}-{b}^{2}}$

Hence, tan ( alpha + beta ) = $\frac{ \tan \alpha + \tan \beta}{1 - \tan \alpha \times \tan \beta}$

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

= ( 2ac / a² - b² )/ (a² - b² - c² + a² /a² - b²)

$\frac{2ac}{{a}^{2}-{c}^{2} }$

Hence proved