Question

for the equation 1-2x-x^2=tan^2(x+y)+cot^2(x+y)​

Answer 1

Step-by-step explanation:

The given equation can be written as

⇒3−2x−x

2

=1+tan

2

(x+y)+1+cot

2

(x+y)

⇒4−(x+1)

2

=sec

2

(x+y)+csc

2

(x+y)

⇒sin

2

(x+y)cos

2

(x+y){4−(x+1)

2

}=1

⇒sin

2

(2x+2y){4−(x+1)

2

}=4 ...(1)

Since sin

2

(2x+2y)≤1 and {4−(x+1)

2

}≤4

Therefore (1) holds only when sin

2

(2x+2y)=1 ...(2)

and {4−(x+1)

2

}=4 ...(3)

From (3), we get x=−1

Putting this in (2), we get sin(2y−2)=±1

⇒2y−2=nπ±

2

π

,n∈Z

⇒y=1+(2n±1)

4

π

,n∈Z

∴x=−1,y=1+(2n±1)

4

π

,n∈Z