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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More