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`        tanA/(1-cotA) + cotA/(1-tanA) -secAcosecA =?`
6 years ago

kunal shrikrishna kasture
33 Points
```										tanA/(1-cotA) + cotA/(1-tanA) - secAcosecA=
(convert all to sine and cosine forms)
we get...
sin(square)A/cosA(sinA - cosA) - cos(square)A/sinA(sinA - cosA) - 1/sinAcosA =
(1/sinA - cosA)(taken as common factor){sin(cube)A - cos(cube)A}/sinAcosA - 1/sinAcosA
using (a^3 - b^3) formula we get
(1 + sinAcosA)/sinAcosA - 1/sinAcosA =
(1 + sinAcosA - 1)/sinAcosA =
sinAcosA/sinAcosA =
1

```
6 years ago
SKS
40 Points
```										The above question can be solved as:
sin a/cos a(1-cos a/sin a) +cos a/sin a(1-sin a/cos a) - 1/cos a*1/sin a
=> sin a/cos a(sin a-cos a/sin a) +cos a/sin a(cos a-sin a/cos a) - 1/sin a.cos a
=>(sin a.sin a)/cos a(sin a-cos a) + (cos a.cos a)/(cos a-sin a) - 1/sin a.cos a
=>(sin a.sin a.sin a) + (-(cos a.cos a.cos a)) -(sin a-cos a)/sin a.cos a(sin a-cos a)
=>(sin a.sin a.sin a) - (cos a.cos a.cos a) - (sin a - cos a)/sin a.cos a(sin a-cos a)
=>[(sin a-cos a)(sin a.sin a +cos a.cos a+sin a.cos a)] -(sin a - cos a)/sin a.cos a(sin a-cos a)
=>(sin a-cos a)[sin a.sin a+cos a.cos a+sin a.cos a - 1]/sin a.cos a(sin a-cos a)
=>(sin a-cos a)[1+sin a.cos a-1]/sin a.cos a(sin a-cos a)
=>(sin a-cos a)(sin a.cos a)/(sin a.cos a)(sin a-cos a)
=>1 ->(ans)
```
6 years ago
rohit kantheti
20 Points
```										write in terms of cosA and sinA.answer is "one"
```
6 years ago
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