Question

Find the value of $\frac{sin{48}^o+cos{42}^o}{cot{42}^o}-\frac{1}{sec{48}^o}$.

• A. $sin{42}^o$
• B. $sin{48}^o$
• C. $sec{48}^o$
• D. $cot{42}^o$

Answer 1

The correct option is A $sin{42}^o$

Consider, $\frac{sin{48}^o+cos{42}^o}{cot{42}^o}-\frac{1}{sec{48}^o}$

$=\frac{sin{48}^o+cos{42}^o}{\frac{cos{42}^o}{sin{42}^o}}-cos{48}^o$

$=\frac{sin{48}^{o}sin{42}^o+sin{42}^{o}cos{42}^o-cos{42}^{o}cos{48}^o}{cos{42}^o}$

$=\frac{-(cos{42}^{o}cos{48}^o-sin{42}^{o}sin{48}^o)+sin{42}^{o}cos{42}^o}{cos{42}^o}$

$=\frac{-cos{90}^o+sin{42}^{o}cos{42}^o}{cos{42}^o}$

$=sin{42}^o$