Sintheta/cottheta+cosectheta =2 + sintheta/cottheta-cosectheta
Answers
Step-by-step explanation:
ANSWER
We have to prove that
cot+cosecθ
sinθ
=2+
cotθ−cosecθ
sinθ
or,
cotθ+cosecθ
sinθ
−
cotθ−cosecθ
sinθ
=2
Now,
⇒ LHS =
cotθ+cosecθ
sinθ
−
cotθ−cosecθ
sinθ
⇒ LHS =
cosecθ+cotθ
sinθ
+
cosecθ−cotθ
sinθ
⇒ LHS = sinθ{
cosecθ+cotθ
1
+
cosecθ−cotθ
1
}
⇒ LHS = $$sin \, \theta\left\{\dfrac{cosec \, \theta \, - \, cot \, \theta \, + \, cosec \, \theta \, + \, cot \, \theta}{cosec^2 \, \theta \, - \, cot^ \, \theta}\right\} \, = \, sin \, \theta \, \left(\dfrac{2 \, cosec \, \theta}{1} \right)$$
⇒ LHS = sinθ(2cosecθ)=2sinθ×
sinθ
1
=2=RHS
⇒ LHS = 2 = RHS
ALTERNANATIVELY,
LHS =
cotθ+cosecθ
sinθ
⇒LHS=sinθ(cosecθ−cotθ) [∵
cosecθ+cotθ
1
=cosecθ−cotθ]
⇒ LHS = sinθ(
sinθ
1
−
sinθ
cosθ
)=sinθ(
sinθ
1−cosθ
)
⇒ LHS = 1 - cos θ
⇒ = 2 - (1 + cos θ)
⇒ LHS = 2 -
1−cosθ
(1+cosθ)(1−cosθ)
⇒ LHS = 2 -
1−cosθ
(1−cos
2
θ)
⇒ LHS = 2 -
1−cosθ
sin
2
θ
=2−
sinθ
1−cosθ
sinθ
=2−
sinθ
1
−
sinθ
cosθ
sinθ
⇒ LHS = 2 -
cosecθ−cotθ
sinθ
= RHS
Step-by-step explanation:
Note : Here I am using A instead of theta.
LHS = sinA/(cotA+cosecA)
= sinA/[(cosA/sinA)+(1/sinA)]
= sinA/[(cosA+1)/sinA]
= sin²A/(1+cosA)
= (1-cos²A)/(1+cosA)
/* sin²A = 1 - cos²A */
= [(1+cosA)(1-cosA)]/(1+cosA)
/* a²-b² = (a+b)(a-b) */
= 1 - cosA ----(1)
RHS = 2+ [sinA/(cotA-cosecA)]
= 2+sinA/[(cosA/sinA)-(1/sinA)]
= 2+SinA/[(cosA-1)/sinA]
= 2+ [ sin²A/(cosA-1)]
= 2 - ( sin²A)/(1-cosA)
= 2- [ (1-cos²A)/(1-cosA)]
= 2-[(1+cosA)(1-cosA)/(1-cosA)]
= 2 - ( 1+cosA)
= 2-1-cosA
= 1-cosA ----(2)
Form (1) & (2) , we conclude that,
(1) = (2)
LHS = RHS
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