11 question pls

11 question pls (O. sin A + cosec A = 2; find the value of sin2 A + cosec2 "f tan A + cot A = 5; find the value of tan2 A + coe A. Given : 4 sin 0 = 3 cos 0; find the (i) sin 0 (ii) cos 0

tan A + cot A = 5 then by squaring on both sides we get,
25
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Arya its not the correct answer
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GPM mein padhte ho
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10th questuon's answer is 2
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(tan A + cot A) = 52                                  [from, (x+y)= x2 + y2 + 2xy]
=  tan2A + cot2A + 2*tan A*cot A = 25
=  tan2A + cot2A = 25 - 2*tan A*    1        ​ [because, tan =   1  ]
                                                       tan A                               cot 
Therefore,
  tan2A + cot2A = 25-2 = 23
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25
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Question 11 plz solution

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tan A + cot A = 5

Squaring both sides
Therefore    (tan A + cot A)2=52
There for    ( tan2 A + cot2 A + 2 x tan2 A x cot​2 A) = 25
Therefore   ( tan A + cot2 A + 2 x tan2 Ax 1/tan2 A) = 25
Therefore    tan2  A  + cot2   A + 2 =25
Threfore     tan2  A   + cot2  A  =  25-2
Therefore    tan2  A  +  cot 2 A =  23
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Test in trigometry
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answer is 25
 
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The answer of Q10 is 2 and of Q11 is 23

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tan A+cot A=5
on squaring both sides
tan2A+cot2A+2.tanA.cotA=25
tan2A+cot2A+2=25
tan2A+cot2A=23
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This is answer

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tan a + cot a=5
tan^a+cot^2a=23
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Answer of 10 is 2
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answer is 246 892
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Tan A + Cot A = 5
(Tan A + Cot A)?. = 5?
Tan? A + Cot?A+ 2?TanA ? CotA. = 25
Tan? A + Cot?A +2?TanA ? 1/TanA=25
Tan? A + Cot?A+2 = 25
Tan? A + Cot?A = 25-2
Tan? A + Cot?A = 23
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