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Question

Prove:
$$sec^{2} \theta + cosec^{2} \theta = sec^{2} \theta \times cosec^{2} \theta$$


Solution

$$\sec^{2}\theta+\csc^{2}\theta=\sec^{2}\theta\times\csc^{2}\theta$$
Given $$\sec^{2}\theta+\csc^{2}\theta$$
We know that
$$\sec\theta=\dfrac{1}{\cos \theta}, \csc\theta=\dfrac{1}{\sin\theta}$$
$$\sec^{2}\theta+\csc^{2}\theta=\dfrac{1}{\cos^{2}\theta}+\dfrac{1}{\sin^{2}\theta}$$
$$\sec^{2}\theta+\csc^{2}\theta=\dfrac{\sin^{2}\theta+\cos^{2}\theta}{\sin^{2}\theta\cos^{2}\theta}$$
$$(\sec^{2}\theta)+(\csc^{2}\theta)=\dfrac{1}{\sin^{2}\theta\cos^{2}\theta} \left[\because \sin^{2}\theta+\cos^{2}\theta=1\right]$$
$$=\dfrac{1}{\sin^{2}\theta}.\dfrac{1}{\cos^{2}\theta}\Rightarrow \dfrac{1}{\cos^{2}\theta}.\dfrac{1}{\sin^{2}\theta}$$
$$(\sec^{2}\theta+\csc^{2}\theta)=\sec^{2}\theta.\csc^{2}\theta\left[\because \sec^{2}\theta=\dfrac{1}{\cos^{2}\theta}\csc^{2}\theta=\dfrac{1}{\sin^{2}\theta}\right]$$

Mathematics

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