is proved
<h3><u>
Solution:</u></h3>
Given that,
------- (1)
First we will simplify the LHS and then compare it with RHS
------ (2)
Substitute this in eqn (2)
On simplification we get,
Cancelling the common terms (sinx + cosx)
We know secant is inverse of cosine
Thus L.H.S = R.H.S
Hence proved
Answer:
a=sqrt(d)-b, a=-sqrt(d)-b
Step-by-step explanation:
(a+b)^2=d
(a+b)=sqrt(d)
a+b=sqrt(d), a+b=-sqrt(d)
a=sqrt(d)-b, a=-sqrt(d)-b
A linear pair of angles has a common side, and the vertex is on the line designated by the points denoting the non-common sides.
Consider the first choice:
... the common sided is RL, and the other two points are P and M. Point R (the vertex of the angles) is not on line PM. (In fact, line PM is not shown on the diagram.)
In the second choice, the given angles do not have a common side. (They are actually "vertical" angles.)
In the 3rd choice, the common side is RN, and vertex point R is found on line MO. These angles <em>are</em> a linear pair:
... - MRN and NRO
