Math, asked by rajusetu, 1 year ago

look at the attachement an dsolve them

Attachments:

rajusetu: you should prove them

Answers

Answered by kvnmurty

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.

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

kvnmurty: click on thanks blue button above pls

rajusetu: yes sir that thanks one i clicked

rajusetu: sir look at commnet i posted plzzzzzzz

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