Trigonometry Help! How do I simplify sec^2/(tan+cot)?

Ok so I have a major assignment due Monday, and I have 20 out of the 26 questions done. I really need help simplifying the following.


A) sec^2/(tan+cot)

B) ((1+sin)/cos)+ (cos/(1+sin))

C) (tan^2/ (1+sec)) cos*sec.

I realize for C that cos and sec reduce to 1 as

:1/cos = sec so cos*(1/cos) = 1 but I cannot get beyond that. Please help! :(

1 Answer

  • Anonymous
    9 years ago
    Favorite Answer

    A) sec^2/(tan+cot)


    (1/cos^2) / ((sin/cos)+(cos/sin))

    (1/cos^2) / ((sin^2 +cos^2)/(cos.sin))

    (1/cos^2) / (1 /(cos.sin)) Since sin^2 + cos^ =1

    (1/cos^2) * (cos.sin)


    tan {ANSWER}

    B) ((1+sin)/cos)+ (cos/(1+sin))

    Taking LCM

    {(1+sin)^2 + cos^2} / {cos(1+sin)}

    Expanding (1+sin)^2 through (a+b)^2 formula

    (1+2sin+sin^2+cos^2) / {cos(1+sin)}

    (1+2sin+1) / {cos(1+sin)} Since sin^2 + cos^ =1

    (2+2sin) / {cos(1+sin)}

    2(1+sin) / {cos(1+sin)}


    2sec ANSWER

    C) (tan^2/ (1+sec)) cos*sec.

    Since cos*sec=1

    (tan^2)/ (1+sec)

    (sec^2-1)/ (1+sec) since 1+tan^2=sec^2

    {(sec+1)(sec-1)} / (1+sec) Expanding (sec^2-1) by (a^2 - b^2) formula

    sec-1 Could be an ANSWER

    Hope this answered !!!

Still have questions? Get your answers by asking now.