Question

d/dx (e^sin2x) diferention dy/dx
​

Answer 1

GIVEN :-

$\\ \sf \: a = 2 + \sqrt{3}$

TO FIND :-

$\\ \sf \: a - \dfrac{1}{a}$

SOLUTION :-

We have,

$\\ \sf \: a = 2 + \sqrt{3}$

Taking reciprocal,

$\\ \sf \: \frac{1}{a} = \frac{1}{2 + \sqrt{3} }$

Rationalising the denominator,

Rationalising factor is 2-√3.

So, multiplying numerator and denominator by 2-√3,

$\\ \implies \sf \: \dfrac{1}{a} = \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \implies \sf \: \dfrac{1}{a} = \dfrac{(2 - \sqrt{3})(2 - \sqrt{3}) }{(2 + \sqrt{3} )(2 - \sqrt{3}) }$

For numerator,

★ (a-b)(a-b) = a² + b² - 2ab

a = 2

b = √3

For denominator,

★ (a+b)(a-b) = a² - b²

a = 2

b = -√3

Putting values,

$\\ \implies \sf \: \dfrac{1}{a} = \dfrac{ {2}^{2} - 2(2)( \sqrt{3}) + {( - \sqrt{3}) }^{2}}{ {2}^{2} - {( - \sqrt{3}) }^{2} } \\ \\ \implies \sf \: \dfrac{1}{a} = \dfrac{4 - 4 \sqrt{3} + 3 }{4 - 3} \\ \\ \implies \sf \: \boxed{ \sf\dfrac{1}{a} = 7 - 4 \sqrt{3} }$

We have ,

$\\ \bullet \: \: \: \: \: \: \sf \: a = 2 + \sqrt{3} \\ \\ \bullet \: \: \: \: \: \sf \: \dfrac{1}{a} = 7 - 4 \sqrt{3}$

So,

$\\ \implies \sf \: a - \dfrac{1}{a} = 2 + {\sqrt{3}} - 7 + { 4 \sqrt{3} } \\ \\ \\ \implies \underbrace{\boxed{ \sf \: a - \dfrac{1}{a } = 8}}$

Answer 2

Answer:

Please find the attached image.