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tanA/(1-cotA) + cotA/(1-tanA) -secAcosecA =?

tanA/(1-cotA) + cotA/(1-tanA) -secAcosecA =?

Grade:10

3 Answers

kunal shrikrishna kasture
33 Points
10 years ago

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

 

SKS
40 Points
10 years ago

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)

rohit kantheti
20 Points
10 years ago
write in terms of cosA and sinA.answer is "one"

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