Question

(sin 47/cos 43°)square
(cos 30°/cot 30°)square - (sin 60°)square ​

Answer 1

Answer:

Step-by-step explanation:-

Cos 43° = sin (90°-43°)=sin 47°

Now

(1)^2 + ( cos 30° × sin 30°/ cos 30° )^3 - (root 3/2)^2

1 + sin 30°^2 - 3/4

1+ 1/4 -3/4

(4+1-3)÷4

=1/2

Answer 2

$(\frac{sin47^{\circ}}{cos43^{\circ}})^2+(\frac{cos30^{\circ}}{cot30^{\circ}})^2-sin^260^{\circ}=\frac{1}{2}$

Step-by-step explanation:

Given:

$(\frac{sin47^{\circ}}{cos43^{\circ}})^2+(\frac{cos30^{\circ}}{cot30^{\circ}})^2-sin^260^{\circ}$

$(\frac{sin47^{\circ}}{cos(90-47)^{\circ}})^2+(\frac{cos30^{\circ}}{cot30^{\circ}})^2-sin^260^{\circ}$

We know that

$cos(90-\theta)=sin\theta$

$cos30^{\circ}=\frac{\sqrt 3}{2}$

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

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

Substitute the values

$(\frac{sin43}{sin43})^2+(\frac{\sqrt 3}{2\times \sqrt 3})^2-(\frac{\sqrt 3}{2})^2$

$1+\frac{1}{4}-\frac{3}{4}$

$1+\frac{1-3}{4}$

$\frac{4-2}{4}=\frac{2}{4}$

$=\frac{1}{2}$

#Learn more:

https://brainly.in/question/2427258:Answered by Dhanvir