Question

cos theta( tan theta/1+sin theta)​

Answer 1

Answer:

L.H.S= secΘ + tan Θ

R.H.S= cosΘ/ 1- sin Θ

taking R.H.S= cosΘ/ 1- sinΘ

multiplying numerator and denominator by (1+ sinΘ)

cosΘ (1+ sinΘ)/ (1- sinΘ) (1 + sinΘ)

cosΘ + cosΘsinΘ / 1 - sin^2Θ

cosΘ + cosΘsinΘ / cos^2 Θ

i.e. cosΘ/ cos^2 + cosΘsinΘ/ cos^2Θ

=> 1 / cosΘ + sinΘ / cosΘ

= secΘ = tan Θ = L.H.S

Answer 2

$cos\theta(\frac{tan\theta}{1 +sin\theta})\\ \\cos\theta(\frac{\frac{sin\theta}{cos\theta} }{1+sin\theta})\\ \\cos\theta(\frac{sin\theta }{cos\theta(1+sin\theta)})\\\\=\frac{sin\theta}{1+sin\theta} \\\\=\frac{sin\theta}{1+sin\theta}*\frac{1-sin\theta}{1-sin\theta}\\ \\=\frac{sin\theta(1-sin\theta)}{1-sin^2\theta} \\\\=\frac{sin\theta(1-sin\theta)}{cos^2\theta}\\ \\=\frac{tan\theta(1-sin\theta)}{cos\theta}$