Question

1+cos^2x=csc^2x what is csc and prove it​

Answer 1

Answer:

this is answer

Step-by-step explanation:

hope it will helpfull

Answer 2

Correction in the question:

Prove 1 + cot²x = csc²x.

What is csc?

Solution:​

⇒ csc is short for cosec.

⇒ cosecθ = hypotenuse/opposite side

Now, let's prove 1 + cot²x = cosec²x. For this purpose, let's draw a right angles triangle with an acute angle x.

In ΔABC,

∠ACB = x

∠B = 90°

Therefore by using Pythagoras theorem we can say that:

$\rm \Longrightarrow AC^2 = AB^2 + BC^2$

Divide both sides by AB².

$\rm \Longrightarrow \dfrac{AC^2}{AB^2} = \dfrac{AB^2}{AB^2} + \dfrac{BC^2}{AB^2}$

$\rm \Longrightarrow \Bigg[\dfrac{AC}{AB}\Bigg]^2 = \Bigg[\dfrac{AB}{AB}\Bigg]^2 + \Bigg[\dfrac{BC}{AB}\Bigg]^2$

We know that:

➝ Hypotenuse/Opposite side = cosecx

➝ Adjacent side/Opposite side = cotx

We also know that:

➝ AB is the side opposite to x.

➝ BC is the side adjacent to x.

➝ AC is the hypotenuse.

Therefore,

➝ AC/AB = Hypotenuse/Opposite side = cosecx

➝ BC/AB = Adjacent Side/Opposite side = cotx

Substitute these values below:

$\rm \Longrightarrow \Bigg[\dfrac{Hypotenuse}{Opposite \ side}\Bigg]^2 = 1 + \Bigg[\dfrac{Adjacent \ side}{Opposite \ side} \Bigg]^2$

$\rm \Longrightarrow \Big[cosecx\Big]^2 = 1 + \Big[cotx\Big]^2$

$\rm \Longrightarrow cosec^2x = 1 + cot^2x$

                or

$\rm \Longrightarrow csc^2x = 1 + cot^2x$

Hence proved.