Results 1 to 2 of 2

Math Help - Need help urently.

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    22

    Need help urently.

    Hi,

    I need to solve below Qs. Can anyone help ASAP.

    1.
    Simplify: tan(180 A) sec(90 + A) cosec (90 + A)
    __________________________________
    Sec (-A) cot (90 + A)


    2. a) sec 120 b) sec 350 c) tan 40

    Please mail me the solution on my mail id, lalitchugh@yahoo.com

    Thanks in advance
    Lalit Chugh
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,751
    Thanks
    651
    Hello, lalitchugh!

    I'll assume you're familiar with the Compound-Angle Identities . . .


    1)\;\text{Simplify: }\:\,\frac{\tan(180-A)\cdot\sec(90+A)\cdot\csc(90+A)}{\sec(-A)\cdot\cot(90+A)}

    I'll simplify each factor first . . .

    \tan(180 - A)\;=\;\frac{\tan180 - \tan A}{1 + \tan180\cdot\tan A} \;= \;\frac{0 - \tan A}{1 + 0\cdot\tan A} \;= \;-\tan A

    \sec(90 + A) \,=\,\frac{1}{\cos(90 + A)} \,=\,\frac{1}{\cos90\cos A - \sin90\sin A} \,= \, \frac{1}{0\cdot\cos A - 1\cdot\sin A} \,= \,\frac{1}{\text{-}\sin A}\,=\,\text{-}\csc A

    \csc(90+A) \,=\,\frac{1}{\sin(90+A)}\,=\,\frac{1}{\sin90\cos A + \cos90\sin A} \,=  \frac{1}{1\cdot\cos A + 0\cdot\sin A} \,= \,\frac{1}{\cos A} \,= \,\sec A

    \sec(-A) \,= \,\sec(A)

    \cot(90 - A)\,=\,\frac{1 + \cos90\cos A}{\cot90 - \cot A} \,= \,\frac{1 + 0\cdot\cot A}{0 - \cot A}\,= \,\frac{1}{\text{-}\cot A}\,=\,-\tan A



    The problem becomes: . \frac{[-\tan A]\,[-\csc A]\,[\sec A]}{[\sec A]\,[-\tan A]} \;=\;-\csc A

    Last edited by Soroban; June 13th 2006 at 08:13 PM.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum