Question

(1-sin theta/1+sin theta)=(sec theta-tan theta)^2​ pls solve fast

Answer 1

Tσ ρяσνє —

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ㅤㅤ$\large\sf{\leadsto \dfrac{1-sin\theta}{1+sin\theta}=(sec\theta-tan\theta)^2 }$

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❍ We know that :

ㅤㅤㅤㅤ$\underline{\boxed{\bf{sin^2 \theta + cos^2 \theta =1 }}}$

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Sσℓνiиg fσя LHS —

ㅤㅤ$\large\sf{\implies \dfrac{1-sin\theta}{1+sin\theta}=(sec\theta-tan\theta)^2 }$

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❍ Multiplying both numerator and denominator by $\large\tt{(1-sin\theta)}$

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ㅤㅤ$\large\sf{\implies \dfrac{(1-sin\theta)(1-sin\theta)}{(1+sin\theta)(1-sin\theta)}}$

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ㅤㅤ$\large\sf{\implies \dfrac{(1-sin\theta)^2}{(1-sin^2 \theta)}}$

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ㅤㅤ$\large\sf{\implies \left(\dfrac{1-sin\theta}{cos\theta}\right)^2}$

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ㅤㅤ$\large\sf{\implies \left(\dfrac{1}{cos\theta}-\dfrac{sin\theta}{cos\theta}\right)^2}$

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ㅤㅤ$\large\sf{\implies (sec\theta-tan\theta)^2}$

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ㅤㅤㅤㅤHєиcє ρяσνєd

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$\bigstar{\large\bf\purple{Identities \:used:}}$

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ㅤㅤ$\large\sf{\longrightarrow (a+b)(a-b)=a^2 -b^2}$

ㅤㅤ$\large\sf{\longrightarrow 1-sin^2 \theta =cos\theta}$

ㅤㅤ$\large\sf{\longrightarrow \dfrac{1}{cos\theta}=sec\theta}$

ㅤㅤ$\large\sf{\longrightarrow \dfrac{sin\theta}{cos\theta}=tan\theta}$

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