To evaluate cot ⁡ 25 + cot ⁡ 55 tan ⁡ 25 + tan ⁡ 55 + cot ⁡ 55 + cot ⁡ 100 tan ⁡ 55 + tan ⁡ 100 + cot ⁡ 100 + cot ⁡ 25 tan ⁡ 100 + tan ⁡ 25

Question

To evaluate             cot      ⁡      25      +      cot      ⁡      55              tan      ⁡      25      +      tan      ⁡      55        +            cot      ⁡      55      +      cot      ⁡      100              tan      ⁡      55      +      tan      ⁡      100        +            cot      ⁡      100      +      cot      ⁡      25              tan      ⁡      100      +      tan      ⁡      25

elpartizano7b3mv · Accepted Answer

Note that 25+55+100=180 so 100=180−(25+55) and   cot  ⁡  (  180  −  θ  )  =  −  cot  ⁡  (  θ  ) so you can rewrite your equation as:  cot  ⁡  (  α  )  cot  ⁡  (  β  )  −  cot  ⁡  (  α  +  β  )  (  cot  ⁡  (  α  )  +  cot  ⁡  (  β  )  )Where   α  =  25 and   β  =  55. Turns out, regardless of the values of   α and   β (as long as cot is defined) the value of the above equation is always 1. Let&#039;s try to prove that.  cot  ⁡  (  α  )  cot  ⁡  (  β  )  =  1  +  cot  ⁡  (  α  +  β  )  (  cot  ⁡  (  α  )  +  cot  ⁡  (  β  )  )Multiplying the whole thing by       sin    2    ⁡  (  α  )  sin  ⁡  (  β  )  cos  ⁡  (  β  )  +  sin  ⁡  (  α  )      sin    2    ⁡  (  β  )  cos  ⁡  (  α  ) yields (after expanding and simplifying):      sin    2    ⁡  (  α  )  sin  ⁡  (  β  )  cos  ⁡  (  β  )  +  sin  ⁡  (  α  )      sin    2    ⁡  (  β  )  cos  ⁡  (  α  )  =      sin    2    ⁡  (  α  )  sin  ⁡  (  β  )  cos  ⁡  (  β  )  +  sin  ⁡  (  α  )      sin    2    ⁡  (  β  )  cos  ⁡  (  α  )Which is indeed true.

Baylee Newman · Accepted Answer

We know that  cot  ⁡  A  +  cot  ⁡  B  =            cot      ⁡      A      cot      ⁡      B      −      1              cot      ⁡      (      A      +      B      )      Now using this identity, we get,  P  =  cot  ⁡      55    ∘    [  cot  ⁡      25    ∘    +  cot  ⁡      100    ∘    ]  +  cot  ⁡      25    ∘    cot  ⁡      100    ∘    P  =  cot  ⁡      55    ∘    [            cot      ⁡              25        ∘            cot      ⁡              100        ∘            −      1              cot      ⁡              125        ∘              ]  +  cot  ⁡      25    ∘    cot  ⁡      100    ∘    P  =  cot  ⁡      25    ∘    cot  ⁡      100    ∘    [  1  +            cot      ⁡              55        ∘                    cot      ⁡              125        ∘              ]  −            cot      ⁡              55        ∘                    cot      ⁡              125        ∘            Now notice   cot  ⁡  (  180  −  α  )  =  −  cot  ⁡  α. When   α  =      55    ∘  , then   cot  ⁡      125    ∘    =  −  cot  ⁡      55    ∘  .Thus,  P  =  cot  ⁡      25    ∘    cot  ⁡      100    ∘    [  1  −  1  ]  −            cot      ⁡              55        ∘                    −      cot      ⁡              55        ∘              =  1

To evaluate cot ⁡ 25 + cot &#x

Answered question

Answer & Explanation

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