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UPSC CDS 2 2022 Math Paper
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© examsnet.com
Question : 50
Total: 100
What is
√
(
1
+
sin
θ
cos
θ
)
2
+
(
cos
θ
1
+
sin
θ
)
2
−
2
equal
to, where
0
,
θ
<
π
2
?
tan
θ
cot
θ
2
tan
θ
2
cot
θ
Validate
Solution:
👈: Video Solution
(
a
−
b
)
2
=
a
2
+
b
2
−
2
a
b
(
a
+
b
)
(
a
−
b
)
=
a
2
−
b
2
sin
θ
cos
θ
=
tan
θ
1
cos
θ
=
sec
θ
Using above formulas:
√
(
1
+
sin
θ
cos
θ
)
2
+
(
cos
θ
1
+
sin
θ
)
2
−
2
⇒
√
(
1
+
sin
θ
cos
θ
)
2
+
(
cos
θ
1
+
sin
θ
)
2
−
2
(
1
+
sin
θ
cos
θ
)
(
cos
θ
1
+
sin
θ
)
⇒
√
(
1
+
sin
θ
cos
θ
−
cos
θ
1
+
sin
θ
)
[
∵
(
a
−
b
)
2
=
a
2
+
b
2
−
2
a
b
]
⇒
(
1
+
sin
θ
cos
θ
−
cos
θ
1
+
sin
θ
)
Since,
sin
θ
∕
cos
θ
=
tan
θ
,
1
∕
cos
θ
=
sec
θ
⇒
(
sec
θ
+
tan
θ
−
1
sec
θ
+
tan
θ
)
⇒
(
sec
θ
+
tan
θ
−
(
sec
θ
−
tan
θ
)
(
sec
θ
+
tan
θ
)
(
sec
θ
−
tan
θ
)
)
⇒
(
sec
θ
+
tan
θ
−
(
sec
θ
−
tan
θ
)
(
sec
2
θ
−
tan
2
θ
)
)
⇒
sec
θ
+
tan
θ
−
sec
θ
+
tan
θ
[
∵
sec
2
θ
−
tan
2
θ
=
1
]
⇒
2
tan
θ
© examsnet.com
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