Question

tanX=b/a then solve equation of a²cosec2x+b²sec2x​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{tan(x)=\dfrac{b}{a}}$

Now,

$\sf{a^2\,cosec(2x)+b^2\,sec(2x)}$

$\sf{=\dfrac{a^2}{sin(2x)}+\dfrac{b^2}{cos(2x)}}$

$\sf{=\dfrac{a^2}{\dfrac{2\,tan(x)}{1+tan^2(x)}}+\dfrac{b^2}{\dfrac{1-tan^2(x)}{1+tan^2(x)}}}$

$\sf{=\dfrac{a^2(1+tan^2(x))}{2\,tan(x)}+\dfrac{b^2(1+tan^2(x))}{1-tan^2(x)}}$

Now, put the value of tan(x)

So,

$\sf{=\dfrac{a^2\bigg\{1+\bigg(\dfrac{b}{a}\bigg)^2\bigg\}}{2\,\bigg(\dfrac{b}{a}\bigg)}+\dfrac{b^2\bigg\{1+\bigg(\dfrac{b}{a}\bigg)^2\bigg\}}{1-\bigg(\dfrac{b}{a}\bigg)^2}}$

$\sf{=\dfrac{a^2\bigg\{\dfrac{a^2+b^2}{a^2}\bigg\}}{\dfrac{2b}{a}}+\dfrac{b^2\bigg\{\dfrac{a^2+b^2}{a^2}\bigg\}}{\dfrac{a^2-b^2}{a^2}}}$

$\sf{=\dfrac{a^2+b^2}{\dfrac{2b}{a}}+\dfrac{b^2(a^2+b^2)}{a^2-b^2}}$

$\sf{=\dfrac{a(a^2+b^2)}{2b}+\dfrac{b^2(a^2+b^2)}{a^2-b^2}}$

$\sf{=(a^2+b^2)\bigg\{\dfrac{a}{2b}+\dfrac{b^2}{a^2-b^2}\bigg\}}$

$\sf{=(a^2+b^2)\bigg\{\dfrac{a(a^2-b^2)+2b^3}{2b(a^2-b^2)}\bigg\}}$

$\sf{=\dfrac{(a^2+b^2)(a^3-ab^2+2b^3)}{2b(a^2-b^2)}}$