Question

if
$\alpha \: and \: \beta$
are the solution of the equation
a tan X + b sec X = c ,
then show that
$\tan( \alpha + \beta ) = \frac{2ac}{ {a}^{2} - {c}^{2} }$
​

Answer 1

Answer:

Refer to the above attachment for your answer!❤️✌️

Answer 2

Step-by-step explanation:

Given that

alpha and beta are solutions of the given

equation

a tan X + b Sec X = c

b sec X = c - a tan X

Now , "Squaring on both sides"

( b sec X )2 = ( c - a tan X )2

b2 sec2 A = c2 + a2 + tan2 X

b2 (1+tan2X) = c2+a2 + tan2 X- 2ac tan X

(b2-a2) tan2 + 2ac tan X + b2-c2 = 0

Let ,

alpha and beta are roots of the solution

==> tan alpha + tan beta = -2ac / b2-a2

==> tan alpha × tan beta = b2-c2 / b2-a2

==> tan(alpha + beta) = tan alpha + tan beta / 1 - tan alpha × tan beta

===> -2ac/b2 - a2 / 1- b2 - c2/b2 - a2

Cancel " b2-a2 " .

Then we get as

===> - 2ac / c2-a2

Inter change the c2 - a2 as " a2 - c2 "

When we change then - 2ac becomes

2ac

===> 2ac / a2 - c2