What is the value of \(\frac{(sin (y-z)\;+ \;sin (y+z)\;+ \;2sin y)}{(sin (x-z)\;+ \;sin (x+z)\;+ \;2sin x)}\):
Answer & explanation
Correct answer: option 4
\(\frac{[sin (y-z)\;+\;sin (y+z)]\;+\;2sin y}{[sin(x-z)\;+\;sin(x+z)]\;+ \;2sin x}\)
= \(\frac{[2sin y\;×\;cos z]\;+\;2sin y}{[2 sin x\;cos z]\;+\;2sin x}\)
= \(\frac{2sin y\;[cos z + 1]}{ 2sin x\;[cos z + 1]}\)
= \(\frac{sin y}{sin x}\)
Alternate method,
Put x = y = z = 45°
Expression becomes ;
\(\frac{sin (45°\;-\;45°)\;+\;sin (45°\;+\;45°)\;+\;2sin 45°}{sin (45°\;-\;45°)\;+\;sin (45°\;+\;45°)\;+\;2sin 45°}\)
= \(\frac{sin 90°\;+\;2sin 45°}{sin 90°\;+\;2sin 45°}\) = 1
⇒ Satisfied by option, (d)
= \(\frac{sin y}{sin x}\) = \(\frac{sin 45°}{sin 45°}\) = 1