Question

cot²x+cot⁴x=cosec⁴x-cosec²x prove it​

Answer 1

||✪✪ QUESTION ✪✪||

prove that cot²x+cot⁴x=cosec⁴x-cosec²x ?

|| ✰✰ ANSWER ✰✰ ||

cot²x+cot⁴x=cosec⁴x-cosec²x can be written as :-

→ cosec⁴x - cot⁴x = cosec²x + cot²x

Now, put cosec(x) = 1/sin(x) and cot(x) = cosx/sinx ,,

→ (1 /sinx)⁴ - (cosx /sinx)⁴ = (1 /sinx)² + (cosx /sinx)²

→ (1 /sin⁴x) - (cos⁴x /sin⁴x) = (1 /sin²x) + (cos²x /sin²x)

→ (1 - cos⁴x) /sin⁴x = (1 + cos²x) /sin²x

Now using (a⁴ - b⁴) = (a² - b²)(a²+b²) in LHS Numerator,

→ [1² - (cos²x)²] /sin⁴x = (1 + cos²x) /sin²x

→ [(1 + cos²x)(1 - cos²x)] /sin⁴x = (1 + cos²x) /sin²x

Now, putting 1 = sin²x + cos²x in LHS, second part Numerator,,

→ {(1 + cos²x) [(sin²x + cos²x) - cos²x]} /sin⁴x = (1 + cos²x) /sin²x

→ [(1 + cos²x) (sin²x + cos²x - cos²x)] /sin⁴x = (1 + cos²x) /sin²x

→ [(1 + cos²x) sin²x] /sin⁴x = (1 + cos²x) /sin²x

→ (1 + cos²x) /sin²x = (1 + cos²x) /sin²x

❦❦ LHS = RHS ❦❦

✪✪ Hence Proved ✪✪

_________________________

✪✪ Short Method ✪✪ :-

→ cosec⁴x - cot⁴x = cosec²x + cot²x

in LHS ,, using (a⁴ - b⁴) = (a² - b²)(a²+b²)

→ (cosec²x - cot²x)(cosec²x + cot²x)

Now, we know that, (cosec²x - cot²x) = 1 ,,

So,

→ 1 * (cosec²x + cot²x)

→ (cosec²x + cot²x) = RHS . (Proved).

_____________________

Answer 2

Answer :

To prove :

• cot²x+cot⁴x = cosec⁴x-cosec²x

Solution :

Rearranging,

cot²x+cosec²x = cosec⁴x-cot⁴x

We know, cosec x =1/sin:x and cot x = cos x/sin x.

Putting these values,

➜ (cosx/sinx)² + (1/sinx)² = (1/sinx)⁴ - (cosx/sinx)⁴

➜ (cos²x + 1)/(sin²x)= (1 - cos⁴x)/(sin⁴x)

➜ (cos²x + 1)/(sin²x) = [(1 - cos²x)²]/(sin⁴x)

➜ (cos²x + 1)/(sin²x) = (1 - cos²x)(1 + cos²x)/(sin⁴x)

✓ Used the identity a² - b² = (a-b)(a+b)

➜ (cos²x + 1)/(sin²x) = {[(sin²x + cos²x) - cos²x][1 + cos²x]}/(sin⁴x)

✓ Used sin²x + cos²x = 1.

➜ (cos²x + 1)/(sin²x) = [(sin²x)(1 + cos²x)]/[sin⁴x]

➜ (cos²x + 1)/(sin²x) = [(1 + cos²x)]/[sin²x]

So, LHS = RHS

$\rule{200}2$

The other method :-

➜ cot²x + cot⁴x = cosec⁴x-cosec²x

➜ cot²x+cosec²x = cosec⁴x-cot⁴x (Rearranging)

➜ cot²x+cosec²x = (cosec²x+cot²x)(cosec²x-cot²x)

[Using a⁴-b⁴ = (a²+b²)(a²-b²)].

➜ cot²x+cosec²x = (cosec²x+cot²x) * 1

[Using identity :- cosec²x - cot²x = 1].

➜ cot²x+cosec²x = cosec²x+cot²x

So, LHS = RHS

$\rule{200}4$