Question

prove that sec square theta + cosec square theta can never be less than 2

Answer 1

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Answer 2

Answer:

We have to prove that:

$\sec^2 \theta+\csc^2 \theta$ can never be less than 2.

Since we know that:

$\sec \theta=\dfrac{1}{\cos \theta}\\\\\\\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

similarly;

$\csc \theta=\dfrac{1}{\sin \theta}\\\\\\\csc^2 \theta=\dfrac{1}{\sin^2 \theta}$

Hence;

$\sec^2 \theta+\csc^2 \theta=\dfrac{1}{\cos^2 \theta}+\dfrac{1}{\sin^2 \theta}\\\\\\\sec^2 \theta+\csc^2 \theta=\dfrac{\sin^2 \theta+\cos^2 \theta}{\sin^2 \theta\cos^2 \theta}\\\\\sec^2 \theta+\csc^2 \theta=\dfrac{1}{\sin^2 \theta\cos^2 \theta}$

( since,

$\sin^2 \theta+\cos^2 \theta=1$

)

Also,

$\sin (2\theta)=2\sin \theta\cos \theta\\\\\sin^2 (2\theta)=4\sin^2 \theta\cos^2 \theta$

Hence,

$\sec^2 \theta+\csc^2 \theta=\dfrac{1}{\dfrac{1}{4}\sin^2 (2\theta)}=\dfrac{4}{\sin^2 (2\theta)}$

Also,

$-1 \leq\sin 2\theta\leq 1\\\\sin^2 (2\theta)\leq 1\\\\\dfrac{1}{\sin^2 (2\theta)}\geq 1\\\\\\\dfrac{4}{\sin^2 (2\theta)}\geq 4$

Hence,

$sec^2 \theta+\csc^2 \theta\geq 4$

Hence it will never be less than 2.