Question

Evaluate. 2tan square 45º+cot square 30º-sin square 60º.​

Answer 1

Answer:

Step-by-step explanation:

To Find :-

$2tan^245^{\circ}+cot^230^{\circ}-sin^260^{\circ}$

Solution :-

We know that :-

$tan45^{\circ}=1$

$cot30^{\circ}=\sqrt{3}$

$sin60^{\circ}=\dfrac{\sqrt{3}}{2}$

$=2(1)^2+(\sqrt{3})^2-\left(\dfrac{\sqrt{3}}{2}\right)^2$

$= 2 + 3 - \dfrac{3}{4}$

$= 5 - \dfrac{3}{4}$

$=\dfrac{20-3}{4}$

$=\dfrac{17}{4}$