1)
LHS = cot(a/2) - tan(a/2)
= (1 - tan^2(a/2))/tan(a/2)
= (2-sec^2(a/2))/tan(a/2)
= 2cot(a/2) - cosec(a/2)sec(a/2)
= 2(1+cos(a))/sin(a) - 1/(cos(a/2)sin(a/2))
= 2 (1+cos(a))/sin(a) - 2/sin(a)) (product to sums)
= 2[(1+cos(a) -1)/sin(a)]
=2cot a
= RHS
2.
LHS = cot(b/2) + tan(b/2)
= [1 + tan^2(b/2)]/tan(b/2)
= sec^2(b/2)/tan(b/2)
= 1/sin(b/2)cos(b/2)
using product to sums
= 2/sin(b)
= 2cosec(b)
= RHS
If George completed 3/5 of work in 9 days, it means that he needs 3 days to finish 1/5th of the work.
The remaining part is 2/5 because 1-3/5=2/5. 2/5 is also 6/15
From this, how much did George do? in the first 3 days he did 1/5th and then one more day was left, during which he did 1/5/3=1/15th.
So he did in total 1/15+1/5=1/15+3/15=4/15.
this means that Paul did the rest of 6/15, so Paul did 6/15-4/15=2/15.
So we know that he does 2/15 in 4 days, which means that every two days he can do 1/15 of the work
so he would need 15 times 2 days to finish the work - 30 days!
Step-by-step explanation:
<h2>Answer:-</h2>
First, we need to add up the x.
We know that 4+8 = 12, so
Now, equating it to y,
Now, if I multiply 
on both sides,
I get,
Now, as 12 got cancelled , I got the final answer as
Hope it helps :)
Answer:
Here you go, I think
Step-by-step explanation:
