Prove the following identities, where the angles involved are acute angles for which the expressions are defined.\( \dfrac{ tan\; θ}{(1-cot\; θ)}+\dfrac{cot\; θ}{(1-tan\; θ)}= 1 + sec\; θ\; cosec\; θ\)

Asked by Abhisek | 1 year ago |  56

1 Answer

Solution :-

L.H.S. = \( \dfrac{ tan \;θ}{(1-cot \;θ)}+ \dfrac{cot \;θ}{(1-tan \;θ)}\)

We know that tan θ =\( \dfrac{ sin\; θ}{cos \;θ}\)

cot θ = \( \dfrac{ cos \;θ } {sin\; θ}\)

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

Answered by Pragya Singh | 1 year ago

Related Questions

Evalute the following:

(i) \( \dfrac{sin 20°}{ cos 70°}\)

(ii) \(\dfrac{ cos 19°}{ sin 71°}\)

(iii) \( \dfrac{sin 21°}{ cos 69°}\)

(iv) \( \dfrac{tan 10°}{ cot 80°}\)

(v) \( \dfrac{sec 11°}{ cosec 79°}\)

Class 10 Maths Introduction to Trigonometry View Answer