Question

Can u please help me in calculating this..


Note : it's multiplication


The answer should be a^2 ...



!!!!!what a shame i couldn't do! ☹️☹️

idk how my teacher solved it​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\dfrac{a(\sqrt{ {b}^{2} - {a}^{2} })cos \theta \: - ab }{ \sqrt{ {a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}} \times \dfrac{ - a(\sqrt{ {b}^{2} - {a}^{2} })cos \theta \: - ab }{ \sqrt{ {a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}}$

$\rm = \:\dfrac{a(\sqrt{ {b}^{2} - {a}^{2} })cos \theta \: - ab }{ \sqrt{ {a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}} \times \dfrac{ - \bigg(a(\sqrt{ {b}^{2} - {a}^{2} })cos \theta + ab \bigg)}{ \sqrt{ {a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}}$

$\rm = \:\dfrac{ab - a(\sqrt{ {b}^{2} - {a}^{2} })cos \theta}{ \sqrt{ {a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}} \times \dfrac{\bigg(a(\sqrt{ {b}^{2} - {a}^{2} })cos \theta + ab \bigg)}{ \sqrt{ {a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}}$

$\rm = \:\dfrac{ab - a(\sqrt{ {b}^{2} - {a}^{2} })cos \theta}{ \sqrt{ {a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}} \times \dfrac{ab + a(\sqrt{ {b}^{2} - {a}^{2} })cos \theta}{ \sqrt{ {a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}}$

We know,

$\: \: \: \: \: \: \: \: \: \red{\bigg \{\:(x + y)(x - y) = {x}^{2} - {y}^{2}\bigg \}}$

So, using this identity, we get

$\rm \: = \: \: \dfrac{ {(ab)}^{2} - \bigg( a(\sqrt{ {b}^{2} - {a}^{2} })cos \theta \bigg)^{2} }{{a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}$

$\rm \: = \: \: \dfrac{ {a}^{2} {b}^{2} - {a}^{2} ( {b}^{2} - {a}^{2})cos^{2} \theta}{{a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}$

$\rm \: = \: \: \dfrac{ {a}^{2} ({b}^{2} - ( {b}^{2} - {a}^{2})cos^{2} \theta)}{{a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}$

$\rm \: = \: \: \dfrac{ {a}^{2} ({b}^{2} - {b}^{2} {cos}^{2}\theta + {a}^{2}cos^{2} \theta)}{{a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}$

$\rm \: = \: \: \dfrac{{a}^{2} \bigg({b}^{2}(1 -{cos}^{2}\theta) + {a}^{2}cos^{2} \theta \bigg)}{{a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}$

$\rm \: = \: \: \dfrac{{a}^{2} \bigg({b}^{2}{sin}^{2}\theta+ {a}^{2}cos^{2} \theta \bigg)}{{a}^{2} {cos}^{2} \theta + {b}^{2} {sin}^{2}\theta}$

$\rm \: = \: \: {a}^{2}$

Hence, Proved

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)