# Look at the attachement an dsolve them

1
by rajusetu
you should prove them

2015-04-14T20:39:53+05:30

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these are all simple exercises,  you need to just substitute the values of trigonometric ratios and simply the expressions.  that is all.

tan² A - Sin² A = Sin²A / Cos² A -  Sin² A
= Sin² A sec²A - Sin²A
= Sin ² A ( sec²A - 1)
= sin² A  tan²A =  sin² A  sin²A  / cos²A
======================
ii
multiply the two terms to get
1+tan A+sec A+cot A+1+cotA SecA - CosecA - cosecA tan A - cosecA secA
= 2 + tanA + secA +cosA/sinA + cosA/sinA *1/cosA - 1/sinA - 1/sinA*sinA/cosA - 1/sinA * 1/cosA
= 2 + sinA/cosA + 1/cosA +CosA/sinA+ 1/sinA -1/sinA - 1/cosA - 1/sinA*1/cosA
= 2 + sinA/cosA + cosA/sinA - 1/sinA * 1/cosA
= 2 + (sin²A +cos²A - 1 ) / (sinA cosA)
=  2
======================
LHS    1/(cosecA- cotA) - 1/sin A
multiply numerator and denominator  with (coseA+ cotA)
(cosecA + cotA)/ 1  - cosec A  = cot A

RHS      1/sinA  -  1/ (cosecA + cotA)
multiply numerator and denominator with  (cosec A - cot A)

cosec A -  (cosecA - cotA) / (cosec²A - cot²A)

=  cot A

LHS = RHS = cot A
========================

(cot A + tan B) / ( cot B + tan A)  =
= (cosA / sinA +  sin B / cosB )    /   (cosB/sinB + sin A/ cosA)
=  (cosA cosB + sinA sin B) * CosA sin B /  [ (cosA cosB + sinA sinB) * sinA cos B ]
= cot A    tan B
as the summation term cancels.

these are all simple exercises, you need to just substitute the values of trigonometric ratios and simply the expressions. that is all.
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