prove that :- tan²theta -cot²theta = (tan theta -cot theta )/(sin theta.cos theta)​ its urgent

Question

prove that :- tan²theta -cot²theta = (tan theta -cot theta )/(sin theta.cos theta)​
its urgent

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1 month 2021-09-17T09:29:27+00:00 1 Answer 0 views 0

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  1. Emma
    0
    2021-09-17T09:30:47+00:00

    To prove :-

    tan²θ – cot²θ = ( tan θ – cot θ ) / ( sin θ . cos θ )

    Proof :-

    Taking LHS

    → LHS =  tan²θ – cot²θ

    putting tan θ = sin θ / cos θ  and cot θ = cos θ / sin θ

    → (sin²θ / cos²θ) – (cos²θ / sin²θ)

    Taking LCM

    → ( (sin²θ)² – (cos²θ)² ) / ( sin²θ . cos²θ )

    using algebraic identity a² – b² = ( a + b ) ( a – b ) in Numerator

    → (( sin²θ + cos²θ ) ( sin²θ – cos²θ )) / ( sin²θ . cos²θ )

    using trigonometric identity

    sin²θ + cos²θ = 1

    → ( 1 ( sin²θ – cos²θ )) / ( sin²θ . cos²θ )

    → ( sin²θ – cos²θ ) / ( sin²θ . cos²θ )

    Now,

    Taking RHS

    → RHS =  ( tan θ – cot θ ) / ( sin θ . cos θ )

    putting tan θ = sin θ / cos θ and cot θ = cos θ / sin θ

    → ( (sin θ / cos θ) – (cos θ / sin θ) ) / ( sin θ . cos θ )

    → ( ( sin²θ – cos²θ ) / ( sin θ . cos θ ) ) / ( sin θ . cos θ )

    → ( sin²θ – cos²θ ) / ( sin²θ . cos²θ )

    so,

    LHS = RHS

    Proved .

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