Prove: (1+ sin 'theta')/cos 'theta' = 1/(sec 'theta' + tan 'theta') | Trigonometric Identity

Question: Prove: $\dfrac{1 + sin \theta}{cos \theta} = \dfrac{1}{sec \theta - tan \theta}$.

Solution:

Taking RHS

$= \dfrac{1}{sec \theta - tan \theta}$

$= \dfrac{1}{\frac{1}{cos \theta} - \frac{sin \theta}{cos \theta}}$

$= \dfrac{1}{\dfrac{1 - sin \theta}{cos \theta}}$

$= 1 × \dfrac{cos \theta}{1 - sin \theta}$

$= \dfrac{cos \theta}{1 - sin \theta}$

$= \dfrac{cos \theta}{1 - sin \theta} × \dfrac{1 + sin \theta}{1 + sin \theta}$

$= \dfrac{cos \theta (1 + sin \theta)}{1 - sin^2 \theta}$

$= \dfrac{cos \theta ( 1 + sin \theta)}{cos^2 \thet}$

$= \dfrac{1 + sin \theta}{cos \theta}$

= LHS

#proved

Related: Prove: $\dfrac{1 - sin \theta}{cos \theta} = \dfrac{1}{sec \theta + tan \theta}$.

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