plzz someone answer my question
Answers
Given, 1/cosec theta + cot theta - 1/sin theta
It can be written as,
= 1/cosec theta + cot theta * cosec theta - cot theta/cosec theta - cot theta - 1/sin theta
= cosec theta - cot theta/cosec^2 theta - cot ^2 theta - 1/sin theta
= cosec theta - cot theta/1 - 1/sin theta
= sin theta(cosec theta - cot theta)-1/sin theta
= 1-cos theta - 1/sin theta
= 1/sin theta - cos theta/sin theta - 1/sin theta
= 1/sin theta - cot theta - cosec theta
= 1/sin theta - (cosec theta + cot theta)
= 1/sin theta - (cosec theta + cot theta) * (cosec theta - cot theta)/cosec theta - cot theta)
= 1/sin theta - 1/(cosec theta - cot theta)
Answer:
Heya...... ✔️✔️✔️
Please consider : theta = x
LHS = 1/(cosec x−cot x) − 1/sin x
= 1/(1/sin x−cosx/sin x) − 1/sinx
= sinx/(1−cosx) − 1/sin x
= sin²x−(1−cosx)/sinx(1−cosx)
= (1−cos²x−1+cosx)/sinx(1−cosx)
......[ ∵ sin²x+cos²x=1]
= cosx(1−cosx)/sinx(1−cosx)
= cot x ......(1)
RHS = 1/sinx − 1/(cosecx+cotx)
= 1/sinx − 1/(1/sinx + cosx/sinx)
= 1/sinx − sinx/(1+cosx)
= (1+cosx−sin²x)/sinx(1+cosx)
= [1+cosx − (1−cos²x)]/sinx(1+cosx)
......[ ∵ sin²x+cos²x=1]
= cosx(1+cosx)/sinx(1+cosx)
= cotx ......(2)
from equations (1) & (2)
∴ LHS = RHS
-------------------------Hence proved –