Question

if alpha and beta are the two solutions of the equation atanx+bsecx=c, then find the values of cos(alpha+beta) and sin (alpha+beta).

Answer 1

Answer:

Step-by-step explanation:

The given equation is:

$atanx+bsecx=c$

⇒$bsecx=c-atanx$

Squaring on both the sides,

$b^{2}sec^{2}x=(c-atanx)^{2}$

$b^{2}sec^{2}x=c^{2}+a^{2}tan^{2}x-2actanx$

$b^{2}(1+tan^{2}x)=c^{2}+a^{2}tan^{2}x-2actanx$

$(b^{2}-a^{2})tan^{2}x+2actanx+b^{2}-c^{2}=0$

Since, α and β are the roots of this equation, therefore

tanα+tanβ=$\frac{-2ac}{b^{2}-c^{2}}$ and

tanαtanβ=$\frac{b^{2}-c^{2}}{b^{2}-a^{2}}$

Now, tan(α+β)=$\frac{tan{\alpha}+tan{\beta}}{1-tan{\alpha}tan{\beta}}$

=$\frac{\frac{-2ac}{b^{2}-a^{2}}}{1-\frac{b^{2}-c^{2}}{b^{2}-a^{2}}}$

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

Then, from figure we get hypotenuse= a+c by using Pythagoras theorem

Sin(α+β)=$\frac{2ac}{a+c}$

cos(α+β)=$\frac{a^{2}-c^{2}}{a+c}$