Prove that, Class 10 Maths Introduction to Trigonometry questions with solutions

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Solution

Q.
Prove that: $\frac{1 + \sec θ - \tan θ}{1 + \sec θ + \tan θ} = \frac{1 - \sin θ}{\cos θ}$

Ans.
$LHS = \frac{1 + \sec θ - \tan θ}{1 + \sec θ + \tan θ}$

$= \frac{\sec^{2} θ - \tan^{2} θ + \sec θ - \tan θ}{1 + \sec θ + \tan θ} = \frac{(\sec θ - \tan θ) (\sec θ + \tan θ + 1)}{1 + \sec θ + \tan θ}$

$= \sec θ - \tan θ = \frac{1}{\cos θ} - \frac{\sin θ}{\cos θ} = \frac{1 - \sin θ}{\cos θ} = RHS$

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