Question

If a sec@+b tan@=c and x and y are the roots of equation find tan(x+y)

Answer 1

✪ANSWER✪

• $\boxed{ tan(x + y) = \frac{2bc}{b^2 - c^2}}$

★GIVEN★

• $a sec(\theta) + b tan(\theta) = c$

• x and y are roots of the given equation.

☆TO FIND☆

• $tan(x+y)$

⍟CALCULATION⍟

$\to a sec(\theta) + b tan(\theta) = c$

$\to a sec(\theta) = c - b tan(\theta)$

$\to a^2sec^2(\theta) = [c - b tan(\theta)]^2$

$\to a^2[1 + tan^2(\theta) = [c - b tan(\theta)]^2$

$\to a^2 + a^2tan^2(\theta)$

$\:\:\:\:\:= c^2 - 2bctan(\theta) + b^2tan^2(\theta)$

$\to (a^2- b^2)tan^2(\theta) + 2bctan(\theta)$

$\:\:\:\:\:+(a^2 - c^2) = 0$

Since x and y are roots of this equation, $tan(x), tan(y)$ are roots of the equation

$\to (a^2- b^2)z^2 + (2bc)z$

$\:\:\:\:\:+(a^2 - c^2) = 0$

So,

$\to tan(x) + tan(y) = - \frac{2bc}{a^2- b^2}$

$\to tan(x)tan(y) = \frac{a^2 - c^2}{a^2- b^2}$

Now we can calculate tan(x+y) as follows,

$\to tan(x + y) = \frac{tan(x) + tan(y)}{1 - tan(x)tan(y)}$

$\to tan(x + y) = \frac{- \frac{2bc}{a^2- b^2}}{1 - \frac{a^2 - c^2}{a^2- b^2}}$

$\to tan(x + y) = \frac{- 2bc}{a^2- b^2-(a^2 - c^2)}$

$\to tan(x + y) = \frac{- 2bc}{- b^2 + c^2}$

$\to tan(x + y) = \frac{2bc}{b^2 - c^2}$