Answer:
The value of given expression [tex]\frac{\csc^2(\frac{pi}{2}-x) }{1+\tan^2 x}[/tex] is 1.
Step-by-step explanation:
Consider the given expression,
[tex]\frac{\csc^2(\frac{pi}{2}-x) }{1+\tan^2 x}[/tex]
We have to simplify to the lowest possible value.
Cosider the expression,
[tex]\frac{\csc^2(\frac{pi}{2}-x) }{1+\tan^2 x}[/tex]
[tex]1+\tan^2 x= \sec^2x[/tex] , we get,
[tex]\Rightarrow \frac{\csc^2(\frac{pi}{2}-x) }{\sec^2x}[/tex]
[tex]\Rightarrow \frac{(\csc(\frac{pi}{2}-x))^2 }{\sec^2x}[/tex]
Also, [tex]\csc(90-x)=\sec x[/tex] , we get,
[tex]\Rightarrow \frac{\sec^2x }{\sec^2x}[/tex]
= 1
Thus, The value of given expression [tex]\frac{\csc^2(\frac{pi}{2}-x) }{1+\tan^2 x}[/tex] is 1.
