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2.Prove the identity cos^2x-sin^2x / cos^2x+sinxcosx =cotx-1/cotx

3.Prove that (sinx+cosx) (tan^2x+1/tanx)=1/cosx +1/sinx

Thanks sooo much and god bless
asked Apr 29, 2013 in TRIGONOMETRY by linda Scholar

2 Answers

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2.

To  prove the identity Presuming cos2x - sin2x / sinxcosx = cotx - 1 / cotx

solve the identity from left hand side

cos2x - sin2x / sinxcosx

Recall : Trigonometry identies cos2x - sin2x = cos2x

cos2x - sin2x / sinxcosx = 2cos2x / 2sinxcosx

Recall : Trigonometry identities 2sinxcosx = sin2x

                                              = 2cos2x / sin2x

Recall : Trigonometry formulas cotA = cosA / sinA

                                              = 2cot2x

Recall : Trigonometry identities cot2x = (cot2x - 1) / 2cotx

                                              = 2( cot2x - 1) / 2cotx

                                             = (cot2x - 1) / cotx

                                             = (cot2x / cotx) - (1 / cotx)

                                             = cotx - (1 / cotx).

answered Apr 29, 2013 by diane Scholar
edited Apr 29, 2013 by diane
0 votes

3.

Solve the identity from left hand side

(sinx + cosx)((tan2x + 1) / tanx) = (sinx + cosx)(((sin2x / cos2x) + 1) / tanx)

                                                         = (sinx + cosx)[(sin2x + cos2x ) / cos2x × tanx]

Recall : Trigonometry identities sin2x + cos2x = 1

                                                        =  (sinx + cosx)[(1) / cos2x × tanx]

Recall : Trigonometry formulas tanx = sinx / cosx

                                                        = (sinx + cosx)[1 / (cos2x × (sinx / cosx))]

                                                       = (sinx + cosx)( 1 / cosx × sinx)

                                                       = sinx / (cosx × sinx) + cosx / (cosx × sinx)

                                                       = (1 / cosx ) + (1 / sinx) .

answered Apr 30, 2013 by diane Scholar

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