Adv
Sonu
  • 0
Newbie

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ [Hint : Write the expression in terms of sin θ and cos θ] Q.5(3)

  • 0

Find out the best solution of the question of  trigonometry of ncert class 10 .Sir please help me to solve the exercise 8.4  question no. 5(3) easily . Its very important question of trigonometry. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ [Hint : Write the expression in terms of sin θ and cos θ]

Share

1 Answer

  1. tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

    L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)

    We know that tan θ =sin θ/cos θ

    cot θ = cos θ/sin θ

    Now, substitute it in the given equation, to convert it in a simplified form

    = [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]

    = [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]

    = sin2θ/[cos θ(sin θ-cos θ)] + cos2θ/[sin θ(cos θ-sin θ)]

    = sin2θ/[cos θ(sin θ-cos θ)] – cos2θ/[sin θ(sin θ-cos θ)]

    = 1/(sin θ-cos θ) [(sin2θ/cos θ) – (cos2θ/sin θ)]

    = 1/(sin θ-cos θ) × [(sin3θ – cos3θ)/sin θ cos θ]

    = [(sin θ-cos θ)(sin2θ+cos2θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]

    = (1 + sin θ cos θ)/sin θ cos θ

    = 1/sin θ cos θ + 1

    = 1 + sec θ cosec θ = R.H.S.

    Therefore, L.H.S. = R.H.S.

    Hence proved

    • 1
Leave an answer

Leave an answer

Browse

Choose from here the video type.

Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs".

Captcha Click on image to update the captcha.

Related Questions