Question

the value of the expression will be? ​

Answer 1

Answer:

if u are in 10th this is out of syllabus

all the best

Answer 2

Answer:

Option (d)

Step-by-step explanation:

Solution :-

Given that

Sin⁶θ + cos⁶θ + 3 sin²θ cos²θ

It can be written as

=> (sin² θ)³ + (cos² θ)³ + 3 sin² θ cos² θ

We know that

a³+b³ = (a+b)³-3ab(a+b)

On applying this identity to the above expression

then

Where a = sin² θ and b = cos² θ

[(sin²θ+cos²θ)³-3sin² θ cos² θ(sin²θ+cos²θ)] +

3 sin² θ cos²θ

=> (1)³ - 3sin² θ cos² θ(1)] + 3 sin² θ cos²θ

Since, sin² θ + cos²θ = 1

=> 1 - 3sin² θ cos² θ+ 3 sin² θ cos²θ

=> 1

Therefore,sin⁶θ+cos⁶θ+3 sin²θ cos²θ = 1

Used formulae:-

→ sin² θ + cos² θ = 1

→ a³+b³ = (a+b)³ - 3ab(a+b)