Prove the following 1- sin A sin (90-1)/cot (90-a)= sin^2 A

Mathematics

1 answer:

Anna [14]2 years ago

7 0

Im going to assume its (90 - A) so we have:-

1 - sinA sin (90 - A) / cot (90 - A)

Convert to sines and cosines:-

= 1 - (sin A cos A) / cos ( 90 - A) / sin ( 90 - A)

= 1 - (sin A cos A) / sin A / cos A

= 1 - cos^2 A

= sin^2 A which is what is required.

You might be interested in

Could someone help me out?

worty [1.4K]

Since the grade of the numerator and the denominator is the same, then the limit exists and is distinct from 0. The limit of the expression is 4/7.

<h3>How to determine the limit of a rational expression when x tends to infinite</h3>

In this problem we must apply some algebraic handling and some known limits to determine whether the limit exists or not. The limit exists if and only if the result exists.

$\lim_{x \to \infty} \frac{4\cdot x - 1}{7\cdot x + 3}$

$\lim_{x \to \infty} \frac{4\cdot x - 1}{7\cdot x + 3} \cdot \frac{x}{x}$

$\lim_{x \to \infty} \frac{4 - \frac{1}{x} }{7 + \frac{3}{x} }$

$\lim_{x \to \infty} \frac{4}{7}$

4/7

Since the grade of the numerator and the denominator is the same, then the limit exists and is distinct from 0. The limit of the expression is 4/7.

To learn more on limits: brainly.com/question/12207558

#SPJ1

8 0

1 year ago

Can somebody help ?

Nuetrik [128]

Answer:

with what

Step-by-step explanation:

8 0

2 years ago

Pls help I will appreciate pls