What is the value of \(\frac{(sin (y-z)\;+ \;sin (y+z)\;+ \;2sin y)}{(sin (x-z)\;+ \;sin (x+z)\;+ \;2sin x)}\):

\(\frac{sin\;y}{sin\;x}\)

\(\frac{[sin (y-z)\;+\;sin (y+z)]\;+\;2sin y}{[sin(x-z)\;+\;sin(x+z)]\;+ \;2sin x}\)

= \(\frac{[2sin y\;×\;cos z]\;+\;2sin y}{[2 sin x\;cos z]\;+\;2sin x}\)

= \(\frac{2sin y\;[cos z + 1]}{ 2sin x\;[cos z + 1]}\)

= \(\frac{sin y}{sin x}\)

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