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]

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# 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)

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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]

= sin

^{2}θ/[cos θ(sin θ-cos θ)] + cos^{2}θ/[sin θ(cos θ-sin θ)]= sin

^{2}θ/[cos θ(sin θ-cos θ)] – cos^{2}θ/[sin θ(sin θ-cos θ)]= 1/(sin θ-cos θ) [(sin

^{2}θ/cos θ) – (cos^{2}θ/sin θ)]= 1/(sin θ-cos θ) × [(sin

^{3}θ – cos^{3}θ)/sin θ cos θ]= [(sin θ-cos θ)(sin

^{2}θ+cos^{2}θ+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