Prove that Root ( sec^2 θ + csc^2 θ) = tan θ + cot θ

Question

Prove that $\sqrt {\sec^2 	heta + \cos ec^2 	heta} = 	an 	heta + \cot 	heta$

Sapling Academy · Accepted Answer

Method 1: We need to prove that $\sqrt {\sec^2 	heta + \cos ec^2 	heta} = 	an 	heta + \cot 	heta$ Let's start with squaring in both sides: $(\sqrt {\sec^2 	heta + \cos ec^2 	heta})^2 = (	an 	heta + \cot 	heta)^2$ $(sec^2 	heta + \cos ec^2 	heta) = 	an ^2 	heta + \cot ^2 	heta + 2 	an 	heta \cot 	heta$ We know that: $ sec^2 	heta = 1 + tan ^2 	heta$ and $\cos ec^2 	heta = 1 + \cot ^2 	heta$ ∴ $(1 + tan ^2 	heta) + (1 + \cot ^2 	heta) = 	an ^2 	heta + \cot ^2 	heta + 2 	an 	heta (\frac{1}{	an 	heta})$ ∴ $(2 + tan ^2 	heta + \cot ^2 	heta) = 	an ^2 	heta + \cot ^2 	heta + 2$ Since LHS = RHS Hence Proved ! Method 2: Let's start from LHS: $\sqrt {\sec^2 	heta + \cos ec^2 	heta}$ Since, $\sec 	heta = \frac{1}{\cos 	heta}$ and $\cos ec 	heta = \frac{1}{\sin 	heta}$ ∴ LHS = $\sqrt {(\frac{1}{\cos 	heta})^2 + (\frac{1}{\sin 	heta})^2}$ = $\sqrt {(\frac{1}{\cos ^2 	heta}) + (\frac{1}{\sin ^2 	heta})}$ = $\sqrt {(\frac{\sin ^2 	heta + \cos ^2 	heta}{\sin ^2 	heta 	imes \cos ^2 	heta)}$ Since, $\sin ^2 	heta + \cos ^2 	heta = 1$ ∴ LHS = $\sqrt {(\frac{1}{\sin ^2 	heta \cos ^2 	heta)}$ = ${\frac{1}{\sin 	heta  \cos 	heta}$ = ${\frac{\sin ^2 	heta + \cos ^2 	heta}{\sin 	heta  \cos 	heta}$   $[\because \sin ^2 	heta + \cos ^2 	heta = 1]$ = ${\frac{\sin ^2 	heta}{\sin 	heta  \cos 	heta} + \frac{\cos ^2 	heta}{\sin 	heta  \cos 	heta}$ = ${\frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta}$ = $	an 	heta + \cot 	heta$ = RHS Hence Proved! Please press the "Heart", if you liked the solution.

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