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

Question

Q) 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!

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