(1-tan A)²+ (1-сot A )² = (SECA-COSECA) ²
Answers
Step-by-step explanation:
Explanation:
(
1
+
tan
A
)
2
+
(
1
+
cot
A
)
2
=
(
sec
A
+
csc
A
)
2
tan
A
=
sin
A
cos
A
cot
A
=
cos
A
sin
A
sec
A
=
1
cos
A
csc
A
=
1
sin
A
(
1
+
sin
A
cos
A
)
2
+
(
1
+
cos
A
sin
A
)
2
=
(
1
cos
A
+
1
sin
A
)
2
(
cos
A
+
sin
A
)
2
cos
2
A
+
(
sin
A
+
cos
A
)
2
sin
2
A
=
(
sin
A
+
cos
A
)
2
sin
A
cos
A
(
cos
A
+
sin
A
)
2
(
1
cos
2
A
+
1
sin
2
A
)
=
(
cos
A
+
sin
A
)
2
(
1
sin
A
cos
A
)
sin
2
A
+
cos
2
A
sin
2
A
cos
2
A
=
1
sin
A
cos
A
1
(
sin
A
cos
A
)
2
=
1
sin
A
cos
A
(
sin
A
cos
A
)
2
=
sin
A
cos
A
(
sin
A
cos
A
)
2
−
sin
A
cos
A
=
0
sin
A
cos
A
(
sin
A
cos
A
−
1
)
=
0
sin
A
=
0
,
cos
A
=
0
,
sin
A
cos
A
−
1
=
0
sin
A
=
0
→
A
=
0
,
π
,
2
π
,
3
π
,
...
cos
A
=
0
→
A
=
π
2
,
3
π
2
,
5
π
2
,
7
π
2
,
...
...
sin
A
cos
A
=
0
→
1
2
sin
2
A
=
0
→
sin
2
A
=
0
2
A
=
0
,
π
,
2
π
,
3
π
,
...
A
=
0
,
π
2
,
π
,
3
π
2
,
2
π
,
5
π
2
,
3
π
,
7
π
2
,
...
...
Hence, the values ofA satisfying the equation
(
1
+
tan
A
)
2
+
(
1
+
cot
A
)
2
=
(
sec
A
+
csc
A
)
2
are,
A
=
...
...
,
−
7
π
2
,
−
3
π
,
−
5
π
2
,
−
2
π
,
−
3
π
2
,
−
π
,
−
π
2
,
0
,
π
2
,
π
,
3
π
2
,
2
π
,
5
π
2
,
3
π
,
7
π
2
,
...
...
Step-by-step explanation:
Given that,
To prove:
(1-tanA)² + (1-cotA)² = (secA-cosecA)²
Consider,LHS
⇒ (1-tanA)² + (1-cotA)²
(∵ (a-b)² = a² + b² - 2ab )
⇒ 1² + tan²A - 2(1)(tanA) + 1² + cot²A -2(1)(cotA)
⇒ (1 + tan²A) + (1 + cot²A) - 2tanA - 2cotA
(∵ sec²A - tan²A = 1 & cosec²A -cot²A = 1 )
⇒ sec²A + cosec²A - 2(tanA + cotA)
⇒ sec²A + cosec²A - 2(sinA/cosA + cosA/sinA)
(∵ tanA = sinA/cosA & cotA = cosA/sinA )
⇒ sec²A + cosec²A - 2((sin²A + cos²A)/(sinA.cosA))
(∵ sin²A + cos²A = 1 )
⇒ sec²A + cosec²A - 2(1/(sinA.cosA)
⇒ sec²A + cosec²A - 2((1/sinA).(1/cosA))
(∵ secA = 1/cosA & cosecA = 1/sinA)
⇒ sec²A + cosec²A - 2(cosecA.secA)
⇒ (secA - cosecA)²
⇒ RHS
∴ LHS = RHS
Hence Proved
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