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Trigonometric Ratios & Identities
NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities
NDA 2018 · Mathematics · Trigonometric Ratios & Identities
If ( (x+y))/( (x-y))=(a+b)/(a-b) , then what is ( x)/( y) equal to?
If
sin
(
x
+
y
)
sin
(
x
−
y
)
=
a
+
b
a
−
b
\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}
s
i
n
(
x
−
y
)
s
i
n
(
x
+
y
)
=
a
−
b
a
+
b
, then what is
tan
x
tan
y
\frac{\tan x}{\tan y}
t
a
n
y
t
a
n
x
equal to?
Options
A
a
b
\frac{\mathrm{a}}{\mathrm{b}}
b
a
B
b
a
\frac{\mathrm{b}}{\mathrm{a}}
a
b
C
a
+
b
a
−
b
\frac{a+b}{a-b}
a
−
b
a
+
b
D
a
−
b
a
+
b
\frac{a-b}{a+b}
a
+
b
a
−
b
Correct answer
A.
a
b
\frac{\mathrm{a}}{\mathrm{b}}
b
a
Step-by-step solution
sin
(
x
+
y
)
sin
(
x
−
y
)
=
a
+
b
a
−
b
\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}
s
i
n
(
x
−
y
)
s
i
n
(
x
+
y
)
=
a
−
b
a
+
b
Applying componendo and dividendo, we get
sin
(
x
+
y
)
+
sin
(
x
−
y
)
sin
(
x
+
y
)
−
sin
(
x
−
y
)
=
a
+
b
+
a
−
b
a
+
b
−
a
+
b
⇒
2
sin
(
x
+
y
+
x
−
y
2
)
cos
(
x
+
y
−
x
+
y
2
)
2
cos
(
x
+
y
+
x
−
y
2
)
sin
(
x
+
y
−
x
+
y
2
)
=
2
a
2
b
⇒
sin
x
⋅
cos
y
cos
x
⋅
sin
y
=
a
b
⇒
tan
x
tan
y
=
a
b
\begin{aligned} & \frac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\frac{a+b+a-b}{a+b-a+b} \\ & \Rightarrow \frac{2 \sin \left(\frac{x+y+x-y}{2}\right) \cos \left(\frac{x+y-x+y}{2}\right)}{2 \cos \left(\frac{x+y+x-y}{2}\right) \sin \left(\frac{x+y-x+y}{2}\right)}=\frac{2 a}{2 b} \\ & \Rightarrow \frac{\sin x \cdot \cos y}{\cos x \cdot \sin y}=\frac{a}{b} \\ & \Rightarrow \frac{\tan x}{\tan y}=\frac{a}{b} \end{aligned}
sin
(
x
+
y
)
−
sin
(
x
−
y
)
sin
(
x
+
y
)
+
sin
(
x
−
y
)
=
a
+
b
−
a
+
b
a
+
b
+
a
−
b
⇒
2
cos
(
2
x
+
y
+
x
−
y
)
sin
(
2
x
+
y
−
x
+
y
)
2
sin
(
2
x
+
y
+
x
−
y
)
cos
(
2
x
+
y
−
x
+
y
)
=
2
b
2
a
⇒
cos
x
⋅
sin
y
sin
x
⋅
cos
y
=
b
a
⇒
tan
y
tan
x
=
b
a
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