Question

tan theta upon 1 minus cot theta + cot theta upon 1 minus 10 theta is equals to 1 + sec theta cosec theta ​

Answer 1

Answer:

As can be seen L.H.S=R.H.S

Answer 2

Hence Proved.

Step-by-step explanation:

Given,

$\frac{tan\theta}{1-cot\theta} +\frac{cot\theta}{1-tan\theta}=1+sec\theta.cosec\theta$

LHS:

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

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

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

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

$=\frac{(sin\theta-cos\theta)(sin^{2}\theta+cos^{2}\theta+sin\theta.cos\theta)}{sin\theta.cos\theta(sin\theta-cos\theta)}$ (Using $x^{3} -y^{3}=(x-y)(x^{2} +y^{2} +xy)$)

$=\frac{(1+sin\theta.cos\theta)}{sin\theta.cos\theta}$ (∵ $sin^{2}\theta+cos^{2}\theta=1$ )

$=1+\frac{1}{sin\theta.cos\theta}$

$=1+sec\theta.cos\theta$ (∵ $\frac{1}{sec\theta}=cos\theta\ and\ \frac{1}{cosec\theta}=sin\theta$ )

$=RHS$

Hence Proved.