Question

(sec theta-1)/sec theta+1 = (cot theta-cosec theta)^2

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$\frac{sec\theta-1}{sec\theta+1}\\=\big(cosec\theta-cot\theta\big)^{2}$

Step-by-step explanation:

$LHS=\frac{sec\theta-1}{sec\theta+1}$

$=\frac{\frac{1}{cos\theta}-1}{\frac{1}{cos\theta}+1}$

$=\frac{\frac{1-cos\theta}{cos\theta}}{\frac{1+cos\theta}{cos\theta}}$

$=\frac{1-cos\theta}{1+cos\theta}$

$=\frac{(1-cos\theta)(1-cos\theta)}{(1+cos\theta)(1-cos\theta)}\\=\frac{(1-cos\theta)^{2}}{1^{2}-cos^{2}\theta}\\=\frac{(1-cos\theta)^{2}}{sin^{2}\theta}\\=\big(\frac{1-cos\theta}{sin\theta}\big)^{2}\\=\big(\frac{1}{sin\theta}-\frac{cos\theta}{sin\theta}\big)^{2}\\=\big(cosec\theta-cot\theta\big)^{2}\\=RHS$

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