<s>[INST] least value of expression What is the least value of $\csc^2(x)+36\sec^2(x)$ ?
So I  differentiated and when  simplifying we get $\tan^4(x)=1/36$, but that is giving me max value i think. Can someone just help me . Or is there any other efficient way. Thanks !
 [/INST] As you stated, yes taking derivative equal to zero you get  $$\tan^4(x)=\frac{1}{36} \rightarrow tan^2(x)=\frac{1}{6}  $$   
More over $$ \frac{1}{\cos^2(x)}= 1+tan^2(x) = 1+ \frac{1}{6} =\frac{7}{6}$$  so the minimum is at such $x$. Substitute in the original formula : 
$$\frac{1}{\sin^2 x} + \frac{36}{\cos^2 x} = \frac{1}{\cos^2(x)} \Big( \frac{cos^2(x)}{\sin^2 (x)} + 36\Big) =\frac{1}{\cos^2(x)} \Big( \frac{1}{\tan^2 (x)} + 36\Big) = \frac{7}{6} ( 6+36) =49 $$
By the way, you may see that also graphically: 

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