Prove that sin θ ( 1 – tan θ ) − cos θ ( 1 − cot θ ) = cosec θ − sec θ

Question

Prove that sin θ ( 1 – tan θ ) − cos θ ( 1 − cot θ ) = cosec θ − sec θ

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Harper 5 months 2021-07-08T02:35:46+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-08T02:36:50+00:00

     \huge \mathcal \colorbox{lightpink}{{Solution:}}

    L:H:S =  \sin0(1 +  \tan0 )  +  \cos0(1 +  \cot0)

     =  \sin0(1   + \frac{ \sin0 }{cos0} ) + cos0(1   + \frac{ \cos0 }{sin0} )

     = (sin0 + cos0)( \frac{sin0}{cos0}  +  \frac{cos0}{sin0} )

     =  \frac{(sin0 + cos0)}{sin0 \: cos0} ( {sin}^{2} 0 +  {cos}^{2}0)

     = ( \frac{1}{cos0}  +  \frac{1}{sin0} )

    sec0 + cosec0 = R:H:S(proved)

    ❥ the above is the answer of your question

    ❥ hope that’s helps you

    0
    2021-07-08T02:36:51+00:00

    Step-by-step explanation:

    L.H.S. = sin θ (1 – tan θ) – cos θ (1- cot θ)

    = sin θ (1 – sin θ/cos θ) – cos θ (1- cos θ/sinθ)

    = sin θ{(cosθ -sinθ )/cos θ} – cos θ{(sinθ-cosθ )/sinθ}

    =(cos θ – sin θ) (sinθ/cos θ – cos θ/sinθ)

    = (cos θ – sin θ)/cos θ sin θ

    = cosec θ – sec θ

    = R.H.S.

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