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Visualisation of Trigonometric Ratios Using a Unit Circle

SecA·√1-sin∧ ...

Question

SecA . √1-sin^2 A

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Solution

√1-sin^2 A = cos A

(√1-sin^2 A) sec A
= cos A sec A
= 1

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\frac{s e c A - t a n A}{s e c A + t a n A} = 1 - 2 s e c A t a n A + 2 t a n^{2} A

\frac{c o t^{2} A (s e c A - 1)}{1 + s i n A} = s e c^{2} A (\frac{1 - s i n A}{1 + s e c A})

(1 + c o t A + t a n A) (s i n A - c o s A)

\frac{s e c A}{c o s e c^{2} A} - \frac{c o s e c A}{s e c^{2} A}

s i n A t a n A - c o t A c o s A

\frac{t a n A}{s e c A - 1} = \frac{t a n A + s e c A + 1}{s e c A - 1 + t a n A}

\frac{{tan} A (cosecA - 1)}{1 + sin A} = secA

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