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

2 thoughts on “Prove that sin θ ( 1 – tan θ ) − cos θ ( 1 − cot θ ) = cosec θ − sec θ”

1. [tex] \huge \mathcal \colorbox{lightpink}{{Solution:}}[/tex]

   [tex]L:H:S = \sin0(1 + \tan0 ) + \cos0(1 + \cot0) [/tex]

   [tex] = \sin0(1 + \frac{ \sin0 }{cos0} ) + cos0(1 + \frac{ \cos0 }{sin0} )[/tex]

   [tex] = (sin0 + cos0)( \frac{sin0}{cos0} + \frac{cos0}{sin0} )[/tex]

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

   [tex] = ( \frac{1}{cos0} + \frac{1}{sin0} )[/tex]

   [tex]sec0 + cosec0 = R:H:S(proved)[/tex]

   ❥ the above is the answer of your question

   ❥ hope that’s helps you

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2. 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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