Sin Cos Tan

[IMQ061000]

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Shouldn't this question be answered like this?

Sin A x 1 / Sin A - (Cos A / (1 / Tan A) = 1 - (Cos A x Tan A)?




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[UCB05]

you forgot an important pair of brackets in your work

 

[IMQ061000]

you forgot an important pair of brackets in your work

I am not understanding how in the explanation they are getting (cos A / sin A) instead of 1/tan A?

 

[lala05]

I am not understanding how in the explanation they are getting (cos A / sin A) instead of 1/tan A?

As UCB05 said, you forgot about the brackets (in red)

 

[UCB05]

um:
cot = 1/tan
tan = sin/cos
therefore 1/tan = 1/(sin/cos) = cos/sin

it's an identity

 

[RCT PC CRN]

I'm not sure whether you got your answer already but here is how I look at the problem like this.

Step 1.

Change everything in sin and cos

sin A [csc A - (cos A/cot A)] = sin A [1/sin A - (sin A)]

---- cot A = cos A/sin A => cos A/cot A = sin A ----

Step 2.

uncover bracket

sin A * 1/sin A - sin A * sin A = 1 - (sin A)^2 = (cos A)^2 - Ans B
--- (sin A)^2 + (cos A)^2 = 1 ------

What I think is in majority of question like this they testing knowledge of basic trig sin vs csc, cos vs sec, tan vs cot and sin2 + cos2 = 1 relation.

 

[IMQ061000]

Thank you all, I really appreciate this. I feel comfortable with the explanation that I receive from SDN. This is very helpful.