the value of x in [-2pi,2pi].for which the graph of function ..y=√1+sinx/1-sinx - secx and y= -√1-sinx /1+sinx+ secx concide are....
in this can we solve by graph???????
if not wht approach shud we use??????/
106are the two equations these?
y=\sqrt{\frac{1+sinx}{1-sinx}}-secx
y=-\sqrt{\frac{1-sinx}{1+sinx}}+secx
106the first equation is y = lsecxl+ltanxl-secx
second equation is y= -lsecxl+ltanxl+secx
they two coincide when lsecxl+ltanxl-secx=-lsecxl+ltanxl+secx
i.e. lsecxl = secx
this is possible when secx is positive i.e.
x= [-2Ï€,-3Ï€/2)U(-Ï€/2,Ï€/2)U(3Ï€/2,2Ï€]
62y=\sqrt{\frac{1+sinx}{1-sinx}}-secx
y=-\sqrt{\frac{1-sinx}{1+sinx}}+secx
their graphs coincide means that they are the same for all values of x.
so this means that
\sqrt{\frac{1+sinx}{1-sinx}}-secx =-\sqrt{\frac{1-sinx}{1+sinx}}+secx
\sqrt{\frac{1+sin x}{1 - sin x}}+\sqrt{\frac{1-sin x}{1+sin x}}=2 sec x\Rightarrow \sqrt{\frac{1+sin x}{1 - sin x}}+\sqrt{\frac{1-sin x}{1+sin x}}=2 sec x \\ \Rightarrow {\frac{|1+sin x|+|1-sin x|}{\sqrt{1 - sin^2 x}}}=2 sec x\\ \Rightarrow {\frac{|1+sin x|+|1-sin x|}{\sqrt{cos^2 x}}}=2 sec x\\ \Rightarrow {\frac{|1+sin x|+|1-sin x|}{|cos x|}}=2 sec x\\ \Rightarrow {|1+sin x|+|1-sin x|}=2\frac{|cos x|}{cos x} \\ \Rightarrow {1+sin x+|1-sin x|}=2\frac{|cos x|}{cos x} \\
The last step is because 1+sin x is positive always..
62also , |1-sin x| is always +ve so you can remove that too
thus cos x = |cos x| is what you have to solve :)
