Prove- tan A- A-1tan A- A-1-1-sin Acos A- A-tan AQ-cos A1-sin A -

#10; (tanA secA) (sec2A tan2A) tanA+1 secA #10; #10; = #10;(tanA secA)(1 (secA tanA)) tanA+1 secA #10; #10; = #10;(tanA secA)(1 sec ...

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Logarithmic Differentiation

Prove: tan A...

Question

Prove:
$\frac{tan A + sec A - 1}{tan A - sec A + 1} = \frac{1 + sin A}{cos A} = sec A + tan A Q = \frac{cos A}{1 - sin A}$ ?

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

\frac{cos A}{1 - tan A} + \frac{sinA}{1 - cot A} = cos A + sinA .

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

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

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

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