What is the minimum value of this expression? $$\sin^2\theta+\cos^2\theta+\csc^2\theta+\sec^2\theta+\tan^2\theta+\cot^2 \theta$$

I tried grouping $\sin^2x+\csc^2x$, $\cos^2x+\sec^2x$ and $\cot^2x+\tan^2x$. I got the answer as $6$ but the book says $7$. How?

marked as duplicate by Blue trigonometry Sep 29 at 12:58

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Since $\tan^2\theta=\frac{1}{\cos^2\theta}-1$ and $\cot^2\theta=\frac{1}{\sin^2\theta}-1$, your expression it's $$2\left(\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}\right)-1=\frac{8}{\sin^22\theta}-1\geq7.$$ The equality occurs for $\theta=45^{\circ}.$

  • What I have done is that I have evaluated $Cos^2x+Sec^2x$ using AM and GM and hence I got the answer as $ 2$ .I have done the same for the other two.Hence I got $2+2+2$ which is $6$ – mampu Sep 29 at 11:52
  • Post please your wrong solution. I'll find a mistake. – Michael Rozenberg Sep 29 at 11:57

$$\sin^2\theta+\cos^2\theta+\csc^2\theta+\sec^2\theta+\tan^2\theta+\cot^2\theta$$

$$=1+1+1+2(\tan^2\theta+\cot^2\theta)$$

Now $\tan^2\theta+\cot^2\theta=(\tan\theta-\cot\theta)^2+2\ge2$

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