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Q) Prove that \sqrt {\sec^2 \theta + \cos ec^2 \theta} = \tan \theta + \cot \theta

Ans:

Method 1: We need to prove that \sqrt {\sec^2 \theta + \cos ec^2 \theta} = \tan \theta + \cot \theta

Let’s start with squaring in both sides:

(\sqrt {\sec^2 \theta + \cos ec^2 \theta})^2 = (\tan \theta + \cot \theta)^2

(sec^2 \theta + \cos ec^2 \theta) = \tan ^2 \theta + \cot ^2 \theta + 2 \tan \theta \cot \theta

We know that:

sec^2 \theta = 1 + tan ^2 \theta and \cos ec^2 \theta = 1 + \cot ^2 \theta

∴ (1 + tan ^2 \theta) + (1 + \cot ^2 \theta) = \tan ^2 \theta + \cot ^2 \theta + 2 \tan \theta (\frac{1}{\tan \theta})

∴ (2 + tan ^2 \theta + \cot ^2 \theta) = \tan ^2 \theta + \cot ^2 \theta + 2

Since LHS = RHS

Hence Proved !

Method 2:

Let’s start from LHS:

\sqrt {\sec^2 \theta + \cos ec^2 \theta}

Since, \sec \theta = \frac{1}{\cos \theta} and \cos ec \theta = \frac{1}{\sin \theta}

∴ LHS = \sqrt {(\frac{1}{\cos \theta})^2 + (\frac{1}{\sin \theta})^2}

= \sqrt {(\frac{1}{\cos ^2 \theta}) + (\frac{1}{\sin ^2 \theta})}

= \sqrt {(\frac{\sin ^2 \theta + \cos ^2 \theta}{\sin ^2 \theta \times \cos ^2 \theta)}

Since, \sin ^2 \theta + \cos ^2 \theta = 1

∴ LHS = \sqrt {(\frac{1}{\sin ^2 \theta \cos ^2 \theta)}

= {\frac{1}{\sin \theta \cos \theta}

= {\frac{\sin ^2 \theta + \cos ^2 \theta}{\sin \theta \cos \theta}   [\because \sin ^2 \theta + \cos ^2 \theta = 1]

= {\frac{\sin ^2 \theta}{\sin \theta \cos \theta} + \frac{\cos ^2 \theta}{\sin \theta \cos \theta}

= {\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}

= \tan \theta + \cot \theta = RHS

Hence Proved!

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