Question

look at teh attachement aND ANSWER THE QUESTIONS

Answer 1

all the following you just have to substitute  sin x/ cos x  = tan x
cos x / sin x = cot x  .  .  and other basic formulas of trigonometric rations. that is all.

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$1-tanA=(cosA-sinA)/CosA\\1-cotA=(sinA-cosA)/sinA\\\\\frac{cosA}{1-tanA}+\frac{sinA}{1-cotA}\\\\=\frac{cos^2A}{cosA-sinA}+\frac{sin^2A}{sinA-cosA}\\\\=\frac{cos^2A-sin^2A}{cosA-sinA}\\\\=(CosA+sinA)(cosA-sinA)/(cosA-sinA)\\\\=cosA+sinA$

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(sinA+ cosA)(cosA/sinA + sinA/cosA)
   =  cosA + sin²A / cosA +  cos²A / sinA + sinA
   =  [Cos²A + sin²A] / cos A + [ cos²A + sin²A ] / sin A
   = 1 / cos A  + 1/sin A    => sec + cosec A
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sec⁴A - sec²A = sec²A (sec²A - 1) = sec²A tan²A = (1+tan²A) tan²A
           = tan²A+ tan⁴A
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cot⁴A + cot²A = cot²A (cot²A + 1) = cot²A cosec²A
   =(cosec²A - 1) cosec²A = cosec⁴ A - cosec²A
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√cosec²A - 1    =  √cot²A = cotA = cos A / sin A = cos A cosec A
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sec²A cosec²A = (1+tan²A)(1+cot²A)
         = 1 + tan²A + cot²A + tan² A cot²A  = answer
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 tan² - sin²  =  sin² / cos² - sin² = sin² [ 1/cos² - 1 ]
          =  sin² (1 - cos² ) / cos²
   =  sin² A   sin² A  sec²A
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