Question

Number of solution(s) of the equation $\tan x=\sin x+1$ in $(-\frac{\pi }{2},\frac{\pi }{2})$ are

• A. 1
• B. 1.00
• C. 01
• D. 1.0

Answer 1

Answer: (A), (B), (C), (D)

Number of solutions of the given equation $\tan x=\sin x+1$ are same as the number of points of intersection of $y=\tan x$ and $y=\sin x+1.$
In the given equation $\tan x=\sin x+1$, fundamental functions used are $\tan x$ and $\sin x$, whose graphs are given by
and
Now we can get the graph of $y=\sin x+1$ by by making a vertical shift by 1 unit to the graph of $y=\sin x$ as shown below
Now we need to check the number of points of intesection in the graph below

There is only one point of intersection.
Hence the given equation has 1 solution in $(-\frac{\pi }{2},\frac{\pi }{2})$