Question

Evaluate with steps: CLASS 10

Answer 1

Answer:

The answer is $\dfrac{13}{8+4\sqrt 3+2\sqrt 2}$

Step-by-step explanation:

   $\displaystyle \frac{5\cos^260^\circ+4\cot^245^\circ-\sec60^\circ}{\cos ec30^\circ+\tan 60^\circ+\sin 45^\circ}$

$=\displaystyle \frac{5(\frac{1}{2})^2+4(1)^2-(2)}{2+\sqrt 3+\frac{1}{\sqrt 2}}$

Using standard angles values of trigonometric ratios

$=\dfrac{\frac{5}{4}+2}{2+\sqrt 3+\frac{1}{\sqrt 2}}=\dfrac{13}{4(2+\sqrt 3+\frac{1}{\sqrt 2})}\\=\dfrac{13}{8+4\sqrt 3+2\sqrt 2}$