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Trigonometric Ratios of Compound Angles

Prove that : ...

Question

Prove that : $\frac{t a n θ + s e c θ - 1}{t a n θ - s e c θ + 1}$ = sec $θ$ + tan $θ$ = $\frac{1 + s i n θ}{c o s θ}$ .

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Solution

LHS = $\frac{t a n θ + s e c θ - 1}{t a n θ - s e c θ + 1}$

LHS = $\frac{(t a n θ + s e c θ) - 1}{(t a n θ - s e c θ) + 1}$

LHS = $\frac{(s e c θ + t a n θ) - (s e c^{2} θ - t a n^{2} θ)}{t a n θ - s e c θ + 1}$

LHS = $\frac{(s e c θ + t a n θ) - (s e c θ + t a n θ) (s e c θ - t a n θ)}{t a n θ - s e c θ + 1}$

LHS = $\frac{(s e c θ + t a n θ) [1 - (s e c θ - t a n θ)]}{t a n θ - s e c θ + 1}$

LHS = $\frac{(s e c θ + t a n θ) (1 - s e c θ + t a n θ)}{t a n θ - s e c θ + 1}$

LHS = $\frac{(s e c θ + t a n θ) (t a n θ - s e c θ + 1)}{(t a n θ - s e c θ + 1)}$

LHS = $s e c θ + t a n θ$

LHS = $\frac{1}{c o s θ} + \frac{s i n θ}{c o s θ}$ = $\frac{1 + s i n θ}{c o s θ}$ = RHS

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Trigonometric Ratios of Compound Angles

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