Question

Solve tan $({45}^0-\frac{A}{2})$ is equal to

• A. $cosecA+cotA$
• B. $cosecA-cotA$
• C. $secA+tanA$
• D. $secA-tanA$

Answer 1

The correct option is D $secA-tanA$

$tan({45}^0-\frac{A}{2})$
$=\frac{1-tan\left(\frac{A}{2}\right)}{1+tan\left(\frac{A}{2}\right)}$
$=\frac{cos\left(\frac{A}{2}\right)-sin\left(\frac{A}{2}\right)}{cos\left(\frac{A}{2}\right)+sin\left(\frac{A}{2}\right)}$
Multiplying and dividing by $\left[cos\right(\frac{A}{2})-sin(\frac{A}{2}\left)\right]$ we get
$=\frac{\left(cos\right(\frac{A}{2})-sin(\frac{A}{2}\left)\right){}^2}{co{s}^2\left(\frac{A}{2}\right)-si{n}^2\left(\frac{A}{2}\right)}$
$=\frac{co{s}^2\left(\frac{A}{2}\right)+si{n}^2\left(\frac{A}{2}\right)-2sin\left(A/2\right)cos\left(A/2\right)}{co{s}^2\left(\frac{A}{2}\right)-si{n}^2\left(\frac{A}{2}\right)}$
$=\frac{1-sinA}{cosA}$
$=\frac{1}{cosA}-\frac{sinA}{cosA}$
$=secA-tanA$