Sin +cos = root 2
Then find
tan +cot=?
Answers
Answered by
3
sin θ + cos θ = √2
we have to find
tanθ + cot θ =?
solution:-
sin θ + cos θ = √2
now square on both side
= ( sin θ + cos θ )² = √2²
= (sin² θ + cos² θ )+ 2 sin θ cos θ = 2
= 1+ 2sin θ cos θ = 2
=> sin θ cos θ = 1/2
now
tanθ + cot θ =sin θ/cos θ + cos θ/sin θ
=( sin² θ +cos² θ) / sin θ cos θ
= 1 / (1/2) = 2 answer
we have to find
tanθ + cot θ =?
solution:-
sin θ + cos θ = √2
now square on both side
= ( sin θ + cos θ )² = √2²
= (sin² θ + cos² θ )+ 2 sin θ cos θ = 2
= 1+ 2sin θ cos θ = 2
=> sin θ cos θ = 1/2
now
tanθ + cot θ =sin θ/cos θ + cos θ/sin θ
=( sin² θ +cos² θ) / sin θ cos θ
= 1 / (1/2) = 2 answer
Answered by
1
Hey !!!
sin + cos = √2
if we squaring on both side we get
(sin + cos )² = (√2)²
=sin² + cos² + 2sin * cos = 2
= 1 + 2sin* cos = 2
= 2sin * cos = 2 -1
= sin * cos = 1/2 ------------1)
now ,
tan + cot = ?
= sin/cos + cos/sin
sin² + cos² /sin* cos
•°• sin² + cos² = 1
so, 1/sin* cos
=1/1/2 [ from equation 1 ]
= 2 Answer
Hope it helps you !!!
#Rajukumar111@@@
sin + cos = √2
if we squaring on both side we get
(sin + cos )² = (√2)²
=sin² + cos² + 2sin * cos = 2
= 1 + 2sin* cos = 2
= 2sin * cos = 2 -1
= sin * cos = 1/2 ------------1)
now ,
tan + cot = ?
= sin/cos + cos/sin
sin² + cos² /sin* cos
•°• sin² + cos² = 1
so, 1/sin* cos
=1/1/2 [ from equation 1 ]
= 2 Answer
Hope it helps you !!!
#Rajukumar111@@@
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