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Math Help - Trigonometry identity proof

  1. #1
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    Trigonometry identity proof

    prove that

    (tan^2 theta)/(1 + tan^2 theta) + (cot^2 theta)/ (1 + cot^2 theta) = 1
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by saha.subham View Post
    prove that

    (tan^2 theta)/(1 + tan^2 theta) + (cot^2 theta)/ (1 + cot^2 theta) = 1
    1 + \tan ^2 a = \sec ^2 a

    1 + \cot ^2 a = \csc ^2 a

    \sec x = \frac{1}{\cos x}

    \csc x = \frac{1}{\sin x}

    \sin ^2 x + \cos ^2 x = 1

    \frac{\tan ^2 \theta }{\sec ^2 \theta } + \frac{\cos ^2 \theta }{\csc ^2 \theta }

    ......continue it
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  3. #3
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    plzz elaborate it , not getting it. i am new to this chapter
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by saha.subham View Post
    prove that

    (tan^2 theta)/(1 + tan^2 theta) + (cot^2 theta)/ (1 + cot^2 theta) = 1
    \frac{\tan ^2 \theta}{1+\tan ^2 \theta} + \frac{\cot ^2 \theta}{1+\cos ^2 \theta}

    \tan x = \frac{\sin x}{\cos x}

    \cot x = \frac{\cos x}{\sin x}

    sub tan and cot values

    \frac{\frac{\sin ^2 x}{\cos ^2 x}}{1 + \frac{\sin ^2 x}{\cos ^2 x}}+\frac{\frac{\cos ^2 x}{\sin ^2 x}}{1+\frac{\cos ^2 x}{\sin ^2 x}}

    in the first fraction multiply both the denominator and nominator by \cos ^2 x and the second fraction by \sin ^2 x

    you will have this

    \frac{\sin ^2 x}{\sin ^2 x+ \cos ^2 x } + \frac{\cos ^2 x}{\cos ^2 x +\sin ^2 x}

    but \sin ^2 x + \cos ^2 x = 1
    so

    \frac{\sin ^2 x }{1} + \frac{\cos ^2 x }{1} = \sin ^2 x + \cos ^2 x = 1 ends
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  5. #5
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    Hello, saha.subham!

    A slightly different approach . . .


    Prove that: . \frac{\tan^2\!\theta}{1 + \tan^2\!\theta}  + \frac{\cot^2\!\theta}{1 + \cot^2\!\theta} \;=\;1

    We have: . \frac{\tan^2\!\theta}{1+\tan^2\!\theta} + \frac{\cot^2\!\theta}{1 + \cot^2\!\theta} \;\;=\;\; \frac{\tan^2\!\theta}{\sec^2\!\theta} + \frac{\cot^2\!\theta}{\csc^2\!\theta} . =\;\;\frac{\dfrac{\sin^2\!\theta}{\cos^2\!\theta}}  {\dfrac{1}{\cos^2\!\theta}} + \frac{\dfrac{\cos^2\!\theta}{\sin^2\!\theta}}{\dfr  ac{1}{\sin^2\!\theta}} \;\;=\;\;\sin^2\!\theta + \cos^2\!\theta \;\;=\;\;1




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