LHS=2t^2/1-t^2
RHS=2t^2/1-t^2
Hence left had side is equal
Step-by-step explanation:
Taking tan x and sec x in the terms of t=tan(x/2)
tan x=2t/1-t^2
sex x=1/cos x
therefore cos x=1-t^2/1+t^2
so sec x=1/1-t^2/1+t^2
sec x=1+t^2/1-t^2
Taking tan x and sec x in the terms of t=tan(x/2)
tan x=2t/1-t^2
sex x=1/cos x
therefore cos x=1-t^2/1+t^2
so sec x=1/1-t^2/1+t^2
sec x=1+t^2/1-t^2
LHS=tanx .tan(x/2)
=2t.t/1-t^2
=2t^2/1-t^2
RHS=sec x-1
=1+t^2/1-t^2-1
=1+t^2-1+t^2/1-t^2
=2t^2/1-t^2
Therefore hence we proved LHS=RHS
=2t.t/1-t^2
=2t^2/1-t^2
RHS=sec x-1
=1+t^2/1-t^2-1
=1+t^2-1+t^2/1-t^2
=2t^2/1-t^2
Therefore hence we proved LHS=RHS
Answer:
D
Step-by-step explanation:
I am pretty sure it is d
A: One solution- a=11 b=5
B: Has no solution- a=10 b=5
C: Infinite solution- a=10 b=-8
Answer:
1. 5.5 ( Cant see numbers that well)
2. 2.1
3. 1.12
4. 0.27
5. 0.84
6. 1.6
Step-by-step explanation:
Line them up with whole numbers and decimals and then subtract right to left.
Answer is D.
As we can observe, in column y
3,4,6,10,18
4-3 = 1
6-4 = 2
10-6 = 4
18-10 = 8
