Question

Prove that identity tan theta upon 1 minus cot theta + cot theta upon 1 minus 10 theta equals to 1 + sec theta cosec theta

Answer 1

check the attachment

Answer 2

Answer:

To show: $\frac{tan\,\theta}{1-cot\,\theta}+\frac{cot\,\theta}{1-tan\,\theta}=cosec\,\theta\:sec\,\theta}+1$

Consider,

$\frac{tan\,\theta}{1-cot\,\theta}+\frac{cot\,\theta}{1-tan\,\theta}$

convert every trigonometric ratio in sin and cos.

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

$=\frac{\frac{sin\,\theta}{cos\,\theta}}{\frac{sin\,\theta-cos\,\theta}{sin\,\theta}}+\frac{\frac{cos\,\theta}{sin\,\theta}}{\frac{cos\,\theta-sin\,\theta}{cos\,\theta}}$

$=\frac{sin^2\,\theta}{cos\,\theta(sin\,\theta-cos\,\theta)}+\frac{cos^2\,\theta}{sin\,\theta(cos\,\theta-sin\,\theta)}$

$=\frac{sin^3\,\theta-cos^3\,\theta}{sin\,\theta\:cos\,\theta(sin\,\theta-cos\,\theta)}$

using, a³ - b³ = ( a - b )( a² + b² + ab )

$=\frac{(sin\,\theta-cos\,\theta)(sin^2\,\theta+cos^2\,\theta+sin\,\theta\:cos\,\theta)}{sin\,\theta\:cos\,\theta(sin\,\theta-cos\,\theta)}$

using sin²x + cos²x = 1

$=\frac{1+sin\,\theta\:cos\,\theta}{sin\,\theta\:cos\,\theta}$

$=\frac{1}{sin\,\theta\:cos\,\theta}+\frac{sin\,\theta\:cos\,\theta}{sin\,\theta\:cos\,\theta}$

$=\frac{1}{sin\,\theta}\times\frac{1}{cos\,\theta}+\frac{sin\,\theta\:cos\,\theta}{sin\,\theta\:cos\,\theta}$

$=cosec\,\theta\:sec\,\theta+1$

$=RHS$

Hence Proved.