e) If $$ \cos \mathrm{A}=\frac{a}{b} $$, prove that $$ \frac{\sin \mathrm{A} \cdot an \mathrm{A}}{\sec \mathrm{A}-1}=\frac{a+b}{b} $$.

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e) If $$ \cos \mathrm{A}=\frac{a}{b} $$, prove that $$ \frac{\sin \mathrm{A} \cdot 	an \mathrm{A}}{\sec \mathrm{A}-1}=\frac{a+b}{b} $$.

UpStudy AI · Accepted Answer

1. Express the original expression in terms of sine and cosine:
   $$
   \frac{\sin \mathrm{A} \cdot \tan \mathrm{A}}{\sec \mathrm{A} - 1} = \frac{\sin \mathrm{A} \cdot \frac{\sin \mathrm{A}}{\cos \mathrm{A}}}{\frac{1}{\cos \mathrm{A}} - 1}
   $$
   
2. Simplify the numerator and denominator:
   $$
   = \frac{\sin^2 \mathrm{A}}{\cos \mathrm{A}\left(\frac{1 - \cos \mathrm{A}}{\cos \mathrm{A}}\right)} = \frac{\sin^2 \mathrm{A}}{1 - \cos \mathrm{A}}
   $$
   
3. Use the Pythagorean identity, $$ \sin^2 \mathrm{A} = 1 - \cos^2 \mathrm{A} $$. Notice that:
   $$
   1 - \cos^2 \mathrm{A} = (1 - \cos \mathrm{A})(1 + \cos \mathrm{A})
   $$
   Substituting into the expression, we have:
   $$
   \frac{(1 - \cos \mathrm{A})(1 + \cos \mathrm{A})}{1 - \cos \mathrm{A}}
   $$
   
4. Cancel the common factor $$ (1 - \cos \mathrm{A}) $$ (assuming $$ \cos \mathrm{A} \neq 1 $$):
   $$
   = 1 + \cos \mathrm{A}
   $$
   
5. Substitute the given value $$ \cos \mathrm{A} = \frac{a}{b} $$:
   $$
   1 + \cos \mathrm{A} = 1 + \frac{a}{b} = \frac{b + a}{b}
   $$
   
Thus, we have proved that:
$$
\frac{\sin \mathrm{A} \cdot \tan \mathrm{A}}{\sec \mathrm{A} - 1} = \frac{a+b}{b}.
$$
$$ \frac{\sin \mathrm{A} \cdot \tan \mathrm{A}}{\sec \mathrm{A} - 1} = \frac{a+b}{b} $$

e) If , prove that .

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