Question

1. (10 points) When verifying identities, common formatting/structure mistakes are: (1) Skipping show- ing the side of the identity you start with; (2) Not including equal signs between steps: (3) Not including the input of the functions (i.e., writing tan instead of tanz): (4) Assuming the identity is true before verifying it (i.e. writing what you are trying to get before you get there) (5) Writing the identity and doing things to both sides, like you were solving an equation. Here are partial verifications of the identity in the box. Each has a has a formatting or structural error. Identify the type of mistake made. (tan’ 1 + 1)(cos-1) = -tan’ I 1.1 (tan-1 +1)(cos1 - 1) = -tan- tan' I cos1 - tan+ cos2-1= -tan'ı tan' I cos' I + cos1-1=-tan'+ tan tan? Icos I+cos I=1 1.4 -tan-1 -(sec-1-1) 1 - sec1 cos 1 sec 1 - sec 1 sect(cos? 1 - 1) MISTAKE: 01 02 03 04 05 MISTAKE: 01 02 03 04 05 1.2 (tan? +1)(cos?-1) = ((sec? -1) + 1)(cos2 -1) = (sec)(cos? -1) 1.5 tan? I cos? I-tan? I + cos21-1 sinº.cos I - tan’ I +cos I-1 cos = sinº I - tanI + cos(z) - 1 = sinº 1 + cos1-tan 1-1 = ... MISTAKE: 01 02 03 04 05 MISTAKE: 01 02 03 04 05 1.3 (tan1 + 1)(cos' 1 - 1) = -tan (tan? 1 + 1)((1 - sin? I) - 1) = - tani (tan1+1)(-sin 1) = - tani MISTAKE: 01 02 03 04 05

Answer 1

The given problem can be solved by identifying the types of error, as directed in the question itself.

A sample proof is also provided in the solution for reference.

To Prove :- tantnt1) (cos'n-i) = -tan'n LHS :- (tan'n ti) (cos 'n-i} = tan'n( cos"n-2)+ 1( cos" n-2) : sec'n-tan'n= 1 ) tan'n = secn-1 2)-(tantn) = 1-sedn (sec'x-a)(cosn-2) + Cos n-1 = Sec'nl costu-2)- 1(cosn- 1) + cos n-1 Sec uxcos'n - sec'n - cosn + 1+ cosn -x %3D = sectu x_I sec'n Sec'n COsn= sec n -) cosn = L Secln 1- sec'n -{ scc'n-2} - { tanta} - - tan'n = RHS > Proved. we will lOOK for erurs in the solution bovided - Foom Now, arert 1.1) the first line of the prooof, R o Stror it is woitten like an equation and Mistake both Solved, hence the sides are foom 1.2) E athe first the function the proof, the input of line of is NOT pvided ie tan instead of tanhe. mistace (3) hence it is waitten Foom 1.3) B the fiost lime of and boot, the sides are Solved the both an equatiun like hence mistave (5 hence mistae NOT used 1.4) The equal to sign in 1.5) In the roof. it is NOT mentioned foom which side the pa0of ib star ted, hence mistake (1)

Please do upvote if you found the solution helpful. Feel free to ask any doubt regarding this question in the comment section. Thanks!