Solve- tan A - sin Atan A - sin A - A - 1 A - 1

tan A + sin Atan A sin A = sin A ( A + 1)sin A ( A 1) = A + 1 A 1 ...

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Circular Measurement of Angle

Solve: tan ...

Question

Solve: $\frac{tan A + sin A}{tan A - sin A} = \frac{sec A + 1}{sec A - 1}$

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Solution

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\sqrt{\frac{1 + sin A}{1 - sin A}} = sec A + tan A

\frac{1 + sin A - cos A}{1 + sin A + cos A} = tan \frac{A}{2}

\frac{tan A + sec A - 1}{tan A - sec A + 1} = sec A + tan A = \frac{1 + sin A}{cos A} = \sqrt{\frac{1 + sin A}{1 - sin A}}

\frac{sin A}{sec A + tan A - 1} + \frac{cos A}{cosec A + cot A - 1} - 1

\frac{tan A + sec A - 1}{tan A - sec A + 1} = \frac{1 + sin A}{cos A} = sec A + tan A Q = \frac{cos A}{1 - sin A}

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