Question

Given: sec A = ${\displaystyle \frac{29}{21},\text{evaluate : sin A}-\frac{1}{\tan \text{A}}}$

Answer 1

Consider the diagram below :

sec A = ${\displaystyle \frac{29}{21}}$

i.e. ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{29}{21}\Rightarrow \frac{\text{AC}}{\text{AB}}=\frac{29}{21}}$

Therefore if length of AB = 21x , length of AC = 29x
Since

AB² + BC² = AC²         ...[ Using Pythagoras Theorem ]

(21x)² + BC² = ( 29x)²

BC² = 841x² – 441x² =400x²

BC = 20x                    ...(perpendicular ) 

Now

sin A = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{20x}{29x}=\frac{20}{29}}$

tan A = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{20x}{21x}=\frac{20}{21}}$

Therefore

sin A – ${\displaystyle \frac{1}{\tan \text{A}}}$

= ${\displaystyle \frac{20}{29}–\frac{1}{\frac{20}{21}}}$

= ${\displaystyle \frac{20}{29}– \frac{21}{20}}$

= – ${\displaystyle \frac{209}{580}}$