Thus L.H.S = R.H.S that is 2/√3cosx + sinx = sec(Π/6-x) is proved
We have to prove that
2/√3cosx + sinx = sec(Π/6-x)
To prove this we will solve the right-hand side of the equation which is
R.H.S = sec(Π/6-x)
= 1/cos(Π/6-x)
[As secƟ = 1/cosƟ)
= 1/[cos Π/6cosx + sin Π/6sinx]
[As cos (X-Y) = cosXcosY + sinXsinY , which is a trigonometry identity where X = Π/6 and Y = x]
= 1/[√3/2cosx + 1/2sinx]
= 1/(√3cosx + sinx]/2
= 2/√3cosx + sinx
R.H.S = L.H.S
Hence 2/√3cosx + sinx = sec(Π/6-x) is proved
Learn more about trigonometry here : brainly.com/question/7331447
#SPJ9
Answer:
General term
=
n
t
h
t
e
r
m
=
a
n
=
5
n
+
2
Explanation:
Here,
7
,
12
,
17
,
22
,
27
,
...
is an Arithmetic Sequence.
The
,
n
t
h
term of Arithmetic Sequence is :
a
n
=
a
1
+
(
n
−
1
)
d
,
w
h
e
r
e
,
a
1
=
first term and
d
=
common difference
We have,
a
1
=
7
and
d
=
12
−
7
=
17
−
12
=
...
=
5
So,
a
n
=
7
+
(
n
−
1
)
⋅
5
a
n
=
7
+
5
n
−
5
a
n
=
5
n
+
2
Hence,
General term
=
n
t
h
t
e
r
m
=
a
n
=
5
n
+
2
Step-by-step explanation:
Answer:
1st Graph:
Add all the numbers (Do the same to 2nd Graph)
Then add
answer will be shown
