Cos A /1-tanA + SinA /1-CotA = COSA + sinA

Question

Cos A
/1-tanA
+
SinA
/1-CotA
=
COSA + sinA

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Remi 2 weeks 2021-09-09T17:54:07+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-09-09T17:55:31+00:00

    Answer:

    Step-by-step explanation:

    \Large{\underline{\underline{\bf{Given:}}}}

    \sf{\dfrac{cos\:A}{1-tan\:A} +\dfrac{sin\:A}{1-cot\:A} =cos\:A+sin\:A}

    \Large{\underline{\underline{\bf{To\:Prove:}}}}

    LHS = RHS

    \Large{\underline{\underline{\bf{Proof:}}}}

    ➜ The LHS is given by

        \sf{\dfrac{cos\:A}{1-tan\:A} +\dfrac{sin\:A}{1-cot\:A}}

    ➜ Simplifying by using identities,

        \sf{\dfrac{cos\:A}{1-\dfrac{sin\:A}{cos\:A} } +\dfrac{sin\:A}{1-\dfrac{cos\:A}{sin\:A} }}

    ➜ Cross multiplying,

        \sf{\dfrac{cos\:A}{\dfrac{cos\:A-sin\:A}{cos\:A} } +\dfrac{sin\:A}{\dfrac{sin\:A-cos\:A}{sin\:A} }}

        \sf{\dfrac{cos^{2} A}{cos\:A-sin\:A}+\dfrac{sin^{2} A}{sin\:A-cos\:A}  }

    ➜ Taking the negative outside,

         \sf{\dfrac{cos^{2} A}{cos\:A-sin\:A}-\dfrac{sin^{2} A}{cos\:A-sin\:A}  }

            \sf{\dfrac{cos^{2}\:A-sin^{2}\:A  }{cos\:A-sin\:A}}

          \sf{\dfrac{(cos\:A+sin\:A)(cos\:A-sin\:A)}{cos\:A-sin\:A}}

    ➜ Cancelling cos A – sin A on both numerator and denominator

         \sf{\dfrac{(cos\:A+sin\:A)\cancel{(cos\:A-sin\:A)}}{\cancel{cos\:A-sin\:A}}}

           \sf{cos\:A+sin\:A}

           = RHS

    ➜ Hence proved.

    \Large{\underline{\underline{\bf{Identities\:used:}}}}

    ➜ tan A = sin A/ cos A

    ➜ cot A = cos A / sin A

    ➜ (a + b) × (a – b) = a² – b²

           

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