Question:

Show that, $\cot \theta + \tan \theta = \csc \theta \times \sec \theta$

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Always use the fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$ to simplify trigonometric expressions.
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Solution and Explanation

L.H.S. $= \cot \theta + \tan \theta$
\[ = \dfrac{\cos \theta}{\sin \theta} + \dfrac{\sin \theta}{\cos \theta} \] \[ = \dfrac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \times \cos \theta} \] \[ = \dfrac{1}{\sin \theta \times \cos \theta} \quad \text{(Since } \sin^2 \theta + \cos^2 \theta = 1) \] \[ = \dfrac{1}{\sin \theta} \times \dfrac{1}{\cos \theta} \] \[ = \csc \theta \times \sec \theta \] Hence, \[ \text{L.H.S.} = \text{R.H.S.} \] \[ \boxed{\therefore \cot \theta + \tan \theta = \csc \theta \times \sec \theta} \]
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